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Codeforces Round 196 (Div. 1) |
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Finished |

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chinese remainder theorem

math

number theory

*2900

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D. GCD Table

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputConsider a table *G* of size *n* × *m* such that *G*(*i*, *j*) = *GCD*(*i*, *j*) for all 1 ≤ *i* ≤ *n*, 1 ≤ *j* ≤ *m*. *GCD*(*a*, *b*) is the greatest common divisor of numbers *a* and *b*.

You have a sequence of positive integer numbers *a*_{1}, *a*_{2}, ..., *a*_{k}. We say that this sequence occurs in table *G* if it coincides with consecutive elements in some row, starting from some position. More formally, such numbers 1 ≤ *i* ≤ *n* and 1 ≤ *j* ≤ *m* - *k* + 1 should exist that *G*(*i*, *j* + *l* - 1) = *a*_{l} for all 1 ≤ *l* ≤ *k*.

Determine if the sequence *a* occurs in table *G*.

Input

The first line contains three space-separated integers *n*, *m* and *k* (1 ≤ *n*, *m* ≤ 10^{12}; 1 ≤ *k* ≤ 10000). The second line contains *k* space-separated integers *a*_{1}, *a*_{2}, ..., *a*_{k} (1 ≤ *a*_{i} ≤ 10^{12}).

Output

Print a single word "YES", if the given sequence occurs in table *G*, otherwise print "NO".

Examples

Input

100 100 5

5 2 1 2 1

Output

YES

Input

100 8 5

5 2 1 2 1

Output

NO

Input

100 100 7

1 2 3 4 5 6 7

Output

NO

Note

Sample 1. The tenth row of table *G* starts from sequence {1, 2, 1, 2, 5, 2, 1, 2, 1, 10}. As you can see, elements from fifth to ninth coincide with sequence *a*.

Sample 2. This time the width of table *G* equals 8. Sequence *a* doesn't occur there.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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