#include <bits/stdc++.h>

const int N = 3054, mod = 1000000007;

int n, q;
int ri[N], f[N][N];
int S[N];
char s[N];

inline int & up(int &x, const int y) {return x < y ? x = y : x;}
inline void add(int &x, const int y) {x += y - mod, x += x >> 31 & mod;}

int main() {
	int i, j, k, l, r, cur, append;
	scanf("%d%d%s", &n, &q, s + 1);
	for (; q; --q) scanf("%d%d", &l, &r), up(ri[l], r);
	for (i = 1; i <= n; ++i) up(up(ri[i], ri[i - 1]), i), S[i] = S[i - 1] + (s[i] & 1);
	f[0][ S[ri[1]] ] = 1, ri[n + 1] = n;
	for (i = 0; i < n; ++i)
		for (j = 0, k = ri[i + 1] - i; k >= 0; ++j, --k) if (cur = f[i][j]) {
			append = S[ri[i + 2]] - S[ri[i + 1]];
			if (j) add(f[i + 1][j - 1 + append], cur);
			if (k) add(f[i + 1][j + append], cur);
		}
	printf("%d\n", *f[n]);
	return 0;
}
### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>

const int N = 3054, mod = 1000000007;

int n, q;
int ri[N], f[N][N];
int S[N];
char s[N];

inline int & up(int &x, const int y) {return x < y ? x = y : x;}
inline void add(int &x, const int y) {x += y - mod, x += x >> 31 & mod;}

int main() {
	int i, j, k, l, r, cur, append;
	scanf("%d%d%s", &n, &q, s + 1);
	for (; q; --q) scanf("%d%d", &l, &r), up(ri[l], r);
	for (i = 1; i <= n; ++i) up(up(ri[i], ri[i - 1]), i), S[i] = S[i - 1] + (s[i] & 1);
	f[0][ S[ri[1]] ] = 1, ri[n + 1] = n;
	for (i = 0; i < n; ++i)
		for (j = 0, k = ri[i + 1] - i; k >= 0; ++j, --k) if (cur = f[i][j]) {
			append = S[ri[i + 2]] - S[ri[i + 1]];
			if (j) add(f[i + 1][j - 1 + append], cur);
			if (k) add(f[i + 1][j + append], cur);
		}
	printf("%d\n", *f[n]);
	return 0;
}
```