#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;
#define rep(i, a, b) for (int i = a; i <= b; ++ i)
const int maxn = 3010;
const int mo = 1000000007;
using namespace std;
int a[maxn], b[maxn], f[maxn][maxn], n, m; char s[maxn];
void solve1(int l, int r) {
	int tot = 0;
	rep(i, l, r) tot += a[i];
	rep(i, l, l + tot - 1) a[i] = 1;
	rep(i, l + tot, r) a[i] = 0;
}
void solve2(int l, int r) {
	int tot = 0;
	rep(i, l, r) tot += 1 - b[i];
	rep(i, l, l + tot - 1) b[i] = 0;
	rep(i, l + tot, r) b[i] = 1;
}
int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	rep(i, 1, n) a[i] = b[i] = (s[i] == '1');
	rep(i, 1, n) b[i] = a[i]; int x, y;
	rep(i, 1, m) scanf("%d%d", &x, &y), solve1(x, y), solve2(x, y);
	rep(i, 2, n) a[i] += a[i - 1], b[i] += b[i - 1];
	f[0][0] = 1;
	rep(i, 1, n) rep(j, b[i], a[i]) f[i][j] = (f[i - 1][j] + f[i - 1][j - 1]) % mo;
	printf("%d\n", f[n][a[n]]);
	return 0;
}
### Prompt

Generate 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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;
#define rep(i, a, b) for (int i = a; i <= b; ++ i)
const int maxn = 3010;
const int mo = 1000000007;
using namespace std;
int a[maxn], b[maxn], f[maxn][maxn], n, m; char s[maxn];
void solve1(int l, int r) {
	int tot = 0;
	rep(i, l, r) tot += a[i];
	rep(i, l, l + tot - 1) a[i] = 1;
	rep(i, l + tot, r) a[i] = 0;
}
void solve2(int l, int r) {
	int tot = 0;
	rep(i, l, r) tot += 1 - b[i];
	rep(i, l, l + tot - 1) b[i] = 0;
	rep(i, l + tot, r) b[i] = 1;
}
int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	rep(i, 1, n) a[i] = b[i] = (s[i] == '1');
	rep(i, 1, n) b[i] = a[i]; int x, y;
	rep(i, 1, m) scanf("%d%d", &x, &y), solve1(x, y), solve2(x, y);
	rep(i, 2, n) a[i] += a[i - 1], b[i] += b[i - 1];
	f[0][0] = 1;
	rep(i, 1, n) rep(j, b[i], a[i]) f[i][j] = (f[i - 1][j] + f[i - 1][j - 1]) % mo;
	printf("%d\n", f[n][a[n]]);
	return 0;
}
```