In this problem, we consider floating-point number formats, data representation formats to approximate real numbers on computers.

Scientific notation is a method to express a number, frequently used for
numbers too large or too small to be written tersely in usual decimal form.
In scientific notation, all numbers are written in the form
*m* × 10^{e}.
Here, *m* (called *significand*) is a number
greater than or equal to 1 and less than 10,
and *e* (called *exponent*) is an integer.
For example, a number 13.5 is equal to 1.35×10^{1},
so we can express it in scientific notation with significand 1.35 and exponent 1.

As binary number representation is convenient on computers,
let's consider *binary scientific notation* with base two, instead of ten.
In binary scientific notation, all numbers are written in the form
*m* × 2^{e}.
Since the base is two, *m* is limited to be less than 2.
For example, 13.5 is equal to 1.6875×2^{3},
so we can express it in binary scientific notation
with significand 1.6875 and exponent 3.
The significand 1.6875 is equal to 1 + 1/2 + 1/8 + 1/16,
which is 1.1011_{2} in binary notation.
Similarly, the exponent 3 can be expressed as 11_{2} in binary notation.

A floating-point number expresses a number in binary scientific notation in finite number of bits.
Although the accuracy of the significand and the range of the exponent are limited by the number of bits, we can express numbers in a wide range with reasonably high accuracy.

In this problem, we consider a 64-bit floating-point number format,
simplified from one actually used widely,
in which only those numbers greater than or equal to 1 can be expressed.
Here, the first 12 bits are used for the exponent and the remaining 52 bits for the significand.
Let's denote the 64 bits of a floating-point number by
*b*_{64}...*b*_{1}.
With *e* an unsigned binary integer
(*b*_{64}...*b*_{53})_{2},
and with *m* a binary fraction represented by the remaining 52 bits
plus one (1.*b*_{52}...*b*_{1})_{2},
the floating-point number represents the number
*m* × 2^{e}.

We show below the bit string of the representation of 13.5 in the format described above.

In floating-point addition operations, the results have to be approximated by numbers representable in floating-point format.
Here, we assume that the approximation is by truncation.
When the sum of two floating-point numbers *a* and *b* is expressed in binary scientific notation as
*a* + *b* = *m* × 2^{e} (1 ≤ *m* < 2, 0 ≤ *e* < 2^{12}),
the result of addition operation on them will be a floating-point number with its first 12 bits representing *e* as an unsigned integer
and the remaining 52 bits representing the first 52 bits of the binary fraction of *m*.

A disadvantage of this approximation method is that the approximation error accumulates easily.
To verify this, let's make an experiment of adding a floating-point number many times, as in the pseudocode shown below.
Here, *s* and *a* are floating-point numbers, and the results of individual addition are approximated as described above.

*s* := *a*
for *n* times {
*s* := *s* + *a*
}

For a given floating-point number *a* and a number of repetitions *n*,
compute the bits of the floating-point number *s* when the above pseudocode finishes.

The input consists of at most 1000 datasets, each in the following format.

*n*

*b*_{52}...*b*_{1}

*n* is the number of repetitions. (1 ≤ *n* ≤ 10^{18})
For each *i*, *b*_{i} is either 0 or 1.
As for the floating-point number *a* in the pseudocode, the exponent is 0 and the significand is *b*_{52}...*b*_{1}.

The end of the input is indicated by a line containing a zero.