Problem: There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

Solution (in CPP):

Solution:
#include <algorithm>
#include <cstdio>
using namespace std;
const int N = 3010;
const int mod = 1e9 + 7;
int n, m;
char s[N];
int f[N][N];
int mxr[N];
int add(int x, int y) {
    x+= y;
    return x >= mod ?x - mod : x;
}
int sub(int x, int y) {
    x -= y;
    return x < 0 ? x +mod : x;
}
int main() {
    scanf("%d %d", &n, &m);
    scanf("%s", s + 1);
    for (int i = 1; i <= n ; i++ )
        mxr[i] = i;
    for (int i = 1; i <= m; i ++) {
        int x, y;
        scanf("%d %d", &x, &y);
        mxr[x] = max(mxr[x], y);
    }
    f[0][0] = 1;
    for (int i = 0; i < n; i ++ ) {
        int ones = 0;
        mxr[i +1] = max(mxr[i+1], mxr[i]);
        if (mxr[i + 1] != mxr[i]) {
            for (int j = mxr[i]+ 1; j <= mxr[i + 1]; j ++)
                if (s[j] == '1')
                    ones ++;
        }
        int all = mxr[i + 1] - i;
        for (int j = 0; j <= n; j++)
            if (f[i][j]) {
                if (ones + j)
                    f[i + 1][ones + j -1] = add(f[i + 1][ones + j - 1], f[i][j]);
                if (ones + j != all)
                    f[i + 1][ones + j] = add(f[i + 1][ones + j], f[i][j]);
            }
    }
    printf("%d\n", f[n][0]);
    return 0;
}