#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
int a[maxn];
char s[maxn];
struct node {
  int f[5][5];
  void init(int x) {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        if (i == j) continue;
        f[i][j] = 0x3f3f3f3f;
      }
    }
    if (x == 2) {
      f[0][0] = 1, f[0][1] = 0;
    } else if (x == 0) {
      f[1][1] = 1, f[1][2] = 0;
    } else if (x == 1) {
      f[2][2] = 1, f[2][3] = 0;
    } else if (x == 7) {
      f[3][3] = 1, f[3][4] = 0;
    } else if (x == 6) {
      f[3][3] = 1;
      f[4][4] = 1;
    } else if (x == -1) {
      for (int i = 0; i < 5; i++) f[i][i] = 0x3f3f3f3f;
    } else {
      for (int i = 0; i < 5; i++) f[i][i] = 0;
    }
  }
  void print() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        if (f[i][j] == 0x3f3f3f3f)
          printf("inf ");
        else
          printf("%d ", f[i][j]);
      }
      printf("\n");
    }
  }
};
node tr[maxn * 4];
node merge(node t1, node t2) {
  node ans;
  ans.init(-1);
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        ans.f[i][j] = min(ans.f[i][j], t1.f[i][k] + t2.f[k][j]);
      }
    }
  }
  return ans;
}
void build(int o, int l, int r) {
  if (l == r) {
    tr[o].init(a[l]);
    return;
  }
  int mid = (l + r) >> 1;
  build((o << 1), l, mid);
  build((o << 1 | 1), mid + 1, r);
  tr[o] = merge(tr[(o << 1)], tr[(o << 1 | 1)]);
}
void update(int o, int l, int r, int ql, int qr, int val) {
  if (l == r) {
    tr[o].init(val);
    return;
  }
  int mid = (l + r) >> 1;
  if (ql <= mid) update((o << 1), l, mid, ql, qr, val);
  if (qr > mid) update((o << 1 | 1), mid + 1, r, ql, qr, val);
  tr[o] = merge(tr[(o << 1)], tr[(o << 1 | 1)]);
}
node query(int o, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) {
    return tr[o];
  }
  int mid = (l + r) >> 1;
  node ans;
  ans.init(-1);
  if (ql <= mid && qr > mid)
    ans = merge(query((o << 1), l, mid, ql, qr),
                query((o << 1 | 1), mid + 1, r, ql, qr));
  else if (ql <= mid)
    ans = query((o << 1), l, mid, ql, qr);
  else if (qr > mid)
    ans = query((o << 1 | 1), mid + 1, r, ql, qr);
  return ans;
}
int main() {
  int n, m, l, r;
  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);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &l, &r);
    node ans = query(1, 1, n, l, r);
    if (ans.f[0][4] == 0x3f3f3f3f)
      printf("-1\n");
    else
      printf("%d\n", ans.f[0][4]);
  }
}

### 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 maxn = 200010;
int a[maxn];
char s[maxn];
struct node {
  int f[5][5];
  void init(int x) {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        if (i == j) continue;
        f[i][j] = 0x3f3f3f3f;
      }
    }
    if (x == 2) {
      f[0][0] = 1, f[0][1] = 0;
    } else if (x == 0) {
      f[1][1] = 1, f[1][2] = 0;
    } else if (x == 1) {
      f[2][2] = 1, f[2][3] = 0;
    } else if (x == 7) {
      f[3][3] = 1, f[3][4] = 0;
    } else if (x == 6) {
      f[3][3] = 1;
      f[4][4] = 1;
    } else if (x == -1) {
      for (int i = 0; i < 5; i++) f[i][i] = 0x3f3f3f3f;
    } else {
      for (int i = 0; i < 5; i++) f[i][i] = 0;
    }
  }
  void print() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        if (f[i][j] == 0x3f3f3f3f)
          printf("inf ");
        else
          printf("%d ", f[i][j]);
      }
      printf("\n");
    }
  }
};
node tr[maxn * 4];
node merge(node t1, node t2) {
  node ans;
  ans.init(-1);
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        ans.f[i][j] = min(ans.f[i][j], t1.f[i][k] + t2.f[k][j]);
      }
    }
  }
  return ans;
}
void build(int o, int l, int r) {
  if (l == r) {
    tr[o].init(a[l]);
    return;
  }
  int mid = (l + r) >> 1;
  build((o << 1), l, mid);
  build((o << 1 | 1), mid + 1, r);
  tr[o] = merge(tr[(o << 1)], tr[(o << 1 | 1)]);
}
void update(int o, int l, int r, int ql, int qr, int val) {
  if (l == r) {
    tr[o].init(val);
    return;
  }
  int mid = (l + r) >> 1;
  if (ql <= mid) update((o << 1), l, mid, ql, qr, val);
  if (qr > mid) update((o << 1 | 1), mid + 1, r, ql, qr, val);
  tr[o] = merge(tr[(o << 1)], tr[(o << 1 | 1)]);
}
node query(int o, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) {
    return tr[o];
  }
  int mid = (l + r) >> 1;
  node ans;
  ans.init(-1);
  if (ql <= mid && qr > mid)
    ans = merge(query((o << 1), l, mid, ql, qr),
                query((o << 1 | 1), mid + 1, r, ql, qr));
  else if (ql <= mid)
    ans = query((o << 1), l, mid, ql, qr);
  else if (qr > mid)
    ans = query((o << 1 | 1), mid + 1, r, ql, qr);
  return ans;
}
int main() {
  int n, m, l, r;
  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);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &l, &r);
    node ans = query(1, 1, n, l, r);
    if (ans.f[0][4] == 0x3f3f3f3f)
      printf("-1\n");
    else
      printf("%d\n", ans.f[0][4]);
  }
}

```