#include <bits/stdc++.h>
using namespace std;
const int INF = 1e6;
int N, Q;
string S;
struct node {
  int m[5][5];
};
node UNIT;
node tree[1 << 19];
node merge(const node& L, const node& R) {
  node x;
  fill(x.m[0], x.m[0] + 25, INF);
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        x.m[i][j] = min(x.m[i][j], L.m[i][k] + R.m[k][j]);
      }
    }
  }
  return x;
}
void init_tree(int k, int l, int r) {
  fill(tree[k].m[0], tree[k].m[0] + 25, INF);
  if (r - l == 1) {
    for (int i = 0; i < 5; i++) {
      tree[k].m[i][i] = 0;
    }
    if (S[l] == '2') {
      tree[k].m[0][1] = 0;
      tree[k].m[0][0] = 1;
    } else if (S[l] == '0') {
      tree[k].m[1][2] = 0;
      tree[k].m[1][1] = 1;
    } else if (S[l] == '1') {
      tree[k].m[2][3] = 0;
      tree[k].m[2][2] = 1;
    } else if (S[l] == '7') {
      tree[k].m[3][4] = 0;
      tree[k].m[3][3] = 1;
    } else if (S[l] == '6') {
      tree[k].m[3][3] = 1;
      tree[k].m[4][4] = 1;
    }
    return;
  }
  int lch = k * 2 + 1;
  int rch = k * 2 + 2;
  int m = (l + r) / 2;
  init_tree(lch, l, m);
  init_tree(rch, m, r);
  tree[k] = merge(tree[lch], tree[rch]);
}
node query(int a, int b, int k, int l, int r) {
  if (b <= l || r <= a) return UNIT;
  if (a <= l && r <= b) return tree[k];
  int m = (l + r) / 2;
  node L = query(a, b, k * 2 + 1, l, m);
  node R = query(a, b, k * 2 + 2, m, r);
  return merge(L, R);
}
int main() {
  int a, b;
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N >> Q;
  cin >> S;
  fill(UNIT.m[0], UNIT.m[0] + 25, INF);
  for (int i = 0; i < 5; i++) UNIT.m[i][i] = 0;
  init_tree(0, 0, N);
  for (int i = 0; i < Q; i++) {
    cin >> a >> b;
    node ans = query(a - 1, b, 0, 0, N);
    cout << (ans.m[0][4] == INF ? -1 : ans.m[0][4]) << endl;
  }
}

### Prompt

Your task is to create a Cpp solution to the following problem:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e6;
int N, Q;
string S;
struct node {
  int m[5][5];
};
node UNIT;
node tree[1 << 19];
node merge(const node& L, const node& R) {
  node x;
  fill(x.m[0], x.m[0] + 25, INF);
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        x.m[i][j] = min(x.m[i][j], L.m[i][k] + R.m[k][j]);
      }
    }
  }
  return x;
}
void init_tree(int k, int l, int r) {
  fill(tree[k].m[0], tree[k].m[0] + 25, INF);
  if (r - l == 1) {
    for (int i = 0; i < 5; i++) {
      tree[k].m[i][i] = 0;
    }
    if (S[l] == '2') {
      tree[k].m[0][1] = 0;
      tree[k].m[0][0] = 1;
    } else if (S[l] == '0') {
      tree[k].m[1][2] = 0;
      tree[k].m[1][1] = 1;
    } else if (S[l] == '1') {
      tree[k].m[2][3] = 0;
      tree[k].m[2][2] = 1;
    } else if (S[l] == '7') {
      tree[k].m[3][4] = 0;
      tree[k].m[3][3] = 1;
    } else if (S[l] == '6') {
      tree[k].m[3][3] = 1;
      tree[k].m[4][4] = 1;
    }
    return;
  }
  int lch = k * 2 + 1;
  int rch = k * 2 + 2;
  int m = (l + r) / 2;
  init_tree(lch, l, m);
  init_tree(rch, m, r);
  tree[k] = merge(tree[lch], tree[rch]);
}
node query(int a, int b, int k, int l, int r) {
  if (b <= l || r <= a) return UNIT;
  if (a <= l && r <= b) return tree[k];
  int m = (l + r) / 2;
  node L = query(a, b, k * 2 + 1, l, m);
  node R = query(a, b, k * 2 + 2, m, r);
  return merge(L, R);
}
int main() {
  int a, b;
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N >> Q;
  cin >> S;
  fill(UNIT.m[0], UNIT.m[0] + 25, INF);
  for (int i = 0; i < 5; i++) UNIT.m[i][i] = 0;
  init_tree(0, 0, N);
  for (int i = 0; i < Q; i++) {
    cin >> a >> b;
    node ans = query(a - 1, b, 0, 0, N);
    cout << (ans.m[0][4] == INF ? -1 : ans.m[0][4]) << endl;
  }
}

```