Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char buf[100000], *p1, *p2;
int rd() {
  int x = 0;
  char s =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (s < '0' || s > '9')
    s = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++);
  while (s >= '0' && s <= '9')
    x = (((x << 2) + x) << 1) + s - '0',
    s = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++);
  return x;
}
namespace SA {
int n, ws[1050], wa[1050], wb[1050], rk[1050], height[1050], sa[1050];
char w[1050];
void build(int m) {
  int i, j, p, *x = wa, *y = wb;
  for (i = 1; i <= m; i++) ws[i] = 0;
  for (i = 1; i <= n; i++) ws[x[i] = w[i]]++;
  for (i = 1; i <= m; i++) ws[i] += ws[i - 1];
  for (i = n; i; i--) sa[ws[x[i]]--] = i;
  for (j = 1; j < n; j <<= 1) {
    for (p = 0, i = n; i > n - j; i--) y[++p] = i;
    for (i = 1; i <= n; i++)
      if (sa[i] > j) y[++p] = sa[i] - j;
    for (i = 1; i <= m; i++) ws[i] = 0;
    for (i = 1; i <= n; i++) ws[x[i]]++;
    for (i = 1; i <= m; i++) ws[i] += ws[i - 1];
    for (i = n; i; i--) sa[ws[x[y[i]]]--] = y[i];
    swap(x, y);
    m = 1;
    for (i = 1; i <= n; i++)
      x[sa[i]] = y[sa[i] + j] == y[sa[i + 1] + j] && y[sa[i]] == y[sa[i + 1]]
                     ? m
                     : m++;
    if (m > n) break;
  }
  for (i = 1; i <= n; i++) rk[sa[i]] = i;
  p = 0;
  for (i = 1; i <= n; i++)
    if (rk[i] != n) {
      j = rk[i] + 1;
      for (; w[i + p] == w[sa[j] + p]; p++)
        ;
      height[rk[i]] = p;
      if (p) p--;
    }
}
void init() {
  n = 6;
  w[1] = 'a';
  w[2] = 'b';
  w[3] = 'a';
  w[4] = 'b';
  w[5] = 'b';
  w[6] = 'a';
  build(128);
}
}  // namespace SA
namespace Scape_Goat_Tree {
int ch[1050][2], val[1050], siz[1050], tot[1050], ex[1050], S[1050], V[1050],
    tp, koishi;
int root, f[1050], cnt;
double AL = 0.75;
void pushup(int p) {
  siz[p] = siz[ch[p][0]] + siz[ch[p][0]] + ex[p];
  tot[p] = tot[ch[p][0]] + tot[ch[p][0]] + 1;
}
int GG(int p) { return tot[p] * AL < max(tot[ch[p][0]], tot[ch[p][0]]); }
int newnode(int v) {
  int p = ++cnt;
  siz[p] = tot[p] = ex[p] = 1;
  val[p] = v;
  return p;
}
void insert(int &p, int v, int fa) {
  if (!p) {
    p = newnode(v);
    f[p] = fa;
    return;
  }
  insert(ch[p][v > val[p]], v, p);
  pushup(p);
  if (GG(p)) koishi = p;
}
void init() {
  insert(root, 1, 0);
  insert(root, 2, 0);
  insert(root, 3, 0);
  insert(root, 4, 0);
  insert(root, 5, 0);
}
}  // namespace Scape_Goat_Tree
int a[100000], n;
int m, b[100050];
long long p[100050];
int main() {
  Scape_Goat_Tree::init();
  SA::init();
  scanf("%d%d", &n, &m);
  int i;
  for (i = 1; i <= n; i++) scanf("%1d", &a[i]);
  for (i = 1; i <= n; i++) {
    if (a[i])
      b[i] = b[i - 1] + 1;
    else
      b[i] = b[i - 1];
  }
  p[0] = 1;
  for (i = 1; i <= n + 1; i++) {
    p[i] = p[i - 1] * 2 % 1000000007;
  }
  int x, y;
  long long ans;
  while (m--) {
    scanf("%d%d", &x, &y);
    int len = y - x + 1, sum = b[y] - b[x - 1];
    if (!sum) {
      printf("0\n");
      continue;
    }
    len -= sum;
    ans = (p[sum] - 1 + (p[sum] - 1) * (p[len] - 1)) % 1000000007;
    printf("%lld\n", (ans % 1000000007 + 1000000007) % 1000000007);
  }
  return 0;
}