#include <bits/stdc++.h>
using namespace std;
const int Maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
struct node {
  int m[5][5];
  node() { memset(m, INF, sizeof(m)); }
};
node tree[4 * Maxn];
char s[Maxn];
string ch = "2017";
node merge(node a, node b) {
  node ret;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k < 5; k++) {
        ret.m[i][j] = min(ret.m[i][j], a.m[i][k] + b.m[k][j]);
      }
    }
  }
  return ret;
}
void build(int root, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 5; i++) tree[root].m[i][i] = 0;
    for (int i = 0; i < 4; i++)
      if (ch[i] == s[l]) {
        tree[root].m[i][i] = 1;
        tree[root].m[i][i + 1] = 0;
        return;
      }
    if (s[l] == '6') {
      tree[root].m[3][3] = 1;
      tree[root].m[4][4] = 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  build(root << 1, l, m);
  build(root << 1 | 1, m + 1, r);
  tree[root] = merge(tree[root << 1], tree[root << 1 | 1]);
}
node query(int root, int l, int r, int L, int R) {
  if (L == l && r == R) return tree[root];
  int m = (l + r) >> 1;
  if (R <= m) return query(root << 1, l, m, L, R);
  if (L > m) return query(root << 1 | 1, m + 1, r, L, R);
  return merge(query(root << 1, l, m, L, m),
               query(root << 1 | 1, m + 1, r, m + 1, R));
}
int main() {
  int n, Q;
  scanf("%d%d", &n, &Q);
  scanf("%s", s + 1);
  build(1, 1, n);
  while (Q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    node ret = query(1, 1, n, l, r);
    if (ret.m[0][4] == INF) {
      printf("-1\n");
      continue;
    }
    printf("%d\n", ret.m[0][4]);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 Maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
struct node {
  int m[5][5];
  node() { memset(m, INF, sizeof(m)); }
};
node tree[4 * Maxn];
char s[Maxn];
string ch = "2017";
node merge(node a, node b) {
  node ret;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      for (int k = 0; k < 5; k++) {
        ret.m[i][j] = min(ret.m[i][j], a.m[i][k] + b.m[k][j]);
      }
    }
  }
  return ret;
}
void build(int root, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 5; i++) tree[root].m[i][i] = 0;
    for (int i = 0; i < 4; i++)
      if (ch[i] == s[l]) {
        tree[root].m[i][i] = 1;
        tree[root].m[i][i + 1] = 0;
        return;
      }
    if (s[l] == '6') {
      tree[root].m[3][3] = 1;
      tree[root].m[4][4] = 1;
    }
    return;
  }
  int m = (l + r) >> 1;
  build(root << 1, l, m);
  build(root << 1 | 1, m + 1, r);
  tree[root] = merge(tree[root << 1], tree[root << 1 | 1]);
}
node query(int root, int l, int r, int L, int R) {
  if (L == l && r == R) return tree[root];
  int m = (l + r) >> 1;
  if (R <= m) return query(root << 1, l, m, L, R);
  if (L > m) return query(root << 1 | 1, m + 1, r, L, R);
  return merge(query(root << 1, l, m, L, m),
               query(root << 1 | 1, m + 1, r, m + 1, R));
}
int main() {
  int n, Q;
  scanf("%d%d", &n, &Q);
  scanf("%s", s + 1);
  build(1, 1, n);
  while (Q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    node ret = query(1, 1, n, l, r);
    if (ret.m[0][4] == INF) {
      printf("-1\n");
      continue;
    }
    printf("%d\n", ret.m[0][4]);
  }
  return 0;
}

```