#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
struct node {
  int f[5][5];
} tr[200005 << 2];
int n, m, a[200005];
node pushup(node x, node y) {
  node ans;
  for (int k = 0; k <= 4; k++) {
    for (int j = 0; j <= 4; j++) {
      if (k == j)
        ans.f[k][j] = 0;
      else
        ans.f[k][j] = inf;
    }
  }
  for (int k = 0; k <= 4; k++) ans.f[k][k] = inf;
  for (int i = 0; i <= 4; ++i)
    for (int j = 0; j <= 4; ++j)
      for (int k = 0; k <= 4; ++k)
        ans.f[i][j] = min(ans.f[i][j], x.f[i][k] + y.f[k][j]);
  return ans;
}
void build(int i, int l, int r) {
  if (l == r) {
    for (int k = 0; k <= 4; k++) {
      for (int j = 0; j <= 4; j++) {
        if (k == j)
          tr[i].f[k][j] = 0;
        else
          tr[i].f[k][j] = inf;
      }
    }
    if (a[l] == -1) {
      for (int k = 0; k <= 4; k++) tr[i].f[k][k] = inf;
    } else if (a[l] == 2) {
      tr[i].f[0][0] = 1;
      tr[i].f[0][1] = 0;
    } else if (a[l] == 0) {
      tr[i].f[1][1] = 1;
      tr[i].f[1][2] = 0;
    } else if (a[l] == 1) {
      tr[i].f[2][2] = 1;
      tr[i].f[2][3] = 0;
    } else if (a[l] == 7) {
      tr[i].f[3][3] = 1;
      tr[i].f[3][4] = 0;
    } else if (a[l] == 6) {
      tr[i].f[3][3] = 1;
      tr[i].f[4][4] = 1;
    }
    return;
  }
  int mid = l + r >> 1;
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  tr[i] = pushup(tr[i * 2], tr[i * 2 + 1]);
}
node query(int i, int l, int r, int x, int y) {
  if (x <= l && r <= y) {
    return tr[i];
  }
  int mid = l + r >> 1;
  if (y <= mid) {
    return query(i * 2, l, mid, x, y);
  }
  if (x > mid) {
    return query(i * 2 + 1, mid + 1, r, x, y);
  }
  return pushup(query(i * 2, l, mid, x, mid),
                query(i * 2 + 1, mid + 1, r, mid + 1, y));
}
char s[200005];
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; ++i) a[i] = s[i] - '0';
  build(1, 1, n);
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    node ans = query(1, 1, n, x, y);
    if (ans.f[0][4] != inf) {
      printf("%d\n", ans.f[0][4]);
    } else {
      printf("-1\n");
    }
  }
  return 0;
}

### Prompt

Generate 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 = 0x3f3f3f3f;
struct node {
  int f[5][5];
} tr[200005 << 2];
int n, m, a[200005];
node pushup(node x, node y) {
  node ans;
  for (int k = 0; k <= 4; k++) {
    for (int j = 0; j <= 4; j++) {
      if (k == j)
        ans.f[k][j] = 0;
      else
        ans.f[k][j] = inf;
    }
  }
  for (int k = 0; k <= 4; k++) ans.f[k][k] = inf;
  for (int i = 0; i <= 4; ++i)
    for (int j = 0; j <= 4; ++j)
      for (int k = 0; k <= 4; ++k)
        ans.f[i][j] = min(ans.f[i][j], x.f[i][k] + y.f[k][j]);
  return ans;
}
void build(int i, int l, int r) {
  if (l == r) {
    for (int k = 0; k <= 4; k++) {
      for (int j = 0; j <= 4; j++) {
        if (k == j)
          tr[i].f[k][j] = 0;
        else
          tr[i].f[k][j] = inf;
      }
    }
    if (a[l] == -1) {
      for (int k = 0; k <= 4; k++) tr[i].f[k][k] = inf;
    } else if (a[l] == 2) {
      tr[i].f[0][0] = 1;
      tr[i].f[0][1] = 0;
    } else if (a[l] == 0) {
      tr[i].f[1][1] = 1;
      tr[i].f[1][2] = 0;
    } else if (a[l] == 1) {
      tr[i].f[2][2] = 1;
      tr[i].f[2][3] = 0;
    } else if (a[l] == 7) {
      tr[i].f[3][3] = 1;
      tr[i].f[3][4] = 0;
    } else if (a[l] == 6) {
      tr[i].f[3][3] = 1;
      tr[i].f[4][4] = 1;
    }
    return;
  }
  int mid = l + r >> 1;
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  tr[i] = pushup(tr[i * 2], tr[i * 2 + 1]);
}
node query(int i, int l, int r, int x, int y) {
  if (x <= l && r <= y) {
    return tr[i];
  }
  int mid = l + r >> 1;
  if (y <= mid) {
    return query(i * 2, l, mid, x, y);
  }
  if (x > mid) {
    return query(i * 2 + 1, mid + 1, r, x, y);
  }
  return pushup(query(i * 2, l, mid, x, mid),
                query(i * 2 + 1, mid + 1, r, mid + 1, y));
}
char s[200005];
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; ++i) a[i] = s[i] - '0';
  build(1, 1, n);
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    node ans = query(1, 1, n, x, y);
    if (ans.f[0][4] != inf) {
      printf("%d\n", ans.f[0][4]);
    } else {
      printf("-1\n");
    }
  }
  return 0;
}

```