#include <bits/stdc++.h>
using namespace std;
int const N = 202020;
char s[N];
int an[N];
int sn[N][10];
struct Node {
  int a[5][5];
  void init() {
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) a[i][j] = 0x3f3f3f3f;
  }
  void init(int l, int r) {
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) a[i][j] = 0x3f3f3f3f;
    if (sn[r][2] - sn[l - 1][2] > 0) {
      a[0][1] = 0;
      a[0][0] = sn[r][2] - sn[l - 1][2];
    } else {
      a[0][0] = 0;
    }
    if (sn[r][0] - sn[l - 1][0] > 0) {
      a[1][2] = 0;
      a[1][1] = sn[r][0] - sn[l - 1][0];
    } else {
      a[1][1] = 0;
    }
    if (sn[r][1] - sn[l - 1][1] > 0) {
      a[2][3] = 0;
      a[2][2] = sn[r][1] - sn[l - 1][1];
    } else {
      a[2][2] = 0;
    }
    if (sn[r][7] - sn[l - 1][7] > 0) {
      a[3][4] = sn[r][6] - sn[l - 1][6];
      a[3][3] = sn[r][6] - sn[l - 1][6] + sn[r][7] - sn[l - 1][7];
    } else {
      a[3][3] = sn[r][6] - sn[l - 1][6];
    }
    a[4][4] = sn[r][6] - sn[l - 1][6];
  }
  void upd(int l, int r, int c) { a[l][r] = min(a[l][r], c); }
} tree[N << 2];
Node merge(Node A, Node B) {
  Node ret;
  ret.init();
  for (int i = 0; i < (5); ++i)
    for (int j = i; j < 5; ++j) {
      for (int k = i; k <= j; ++k) {
        int lt = A.a[i][k];
        int rt = B.a[k][j];
        ret.upd(i, j, lt + rt);
      }
    }
  return ret;
}
void build(int l, int r, int p) {
  if (l == r) {
    tree[p].init(l, l);
    return;
  }
  int mid = (l + r) >> 1, lp = p << 1, rp = p << 1 | 1;
  build(l, mid, lp);
  build(mid + 1, r, rp);
  tree[p] = merge(tree[lp], tree[rp]);
}
Node qry(int l, int r, int a, int b, int p) {
  if (l == a && r == b) {
    return tree[p];
  }
  int mid = (l + r) >> 1, lp = p << 1, rp = p << 1 | 1;
  if (b <= mid)
    return qry(l, mid, a, b, lp);
  else if (a > mid)
    return qry(mid + 1, r, a, b, rp);
  else {
    return merge(qry(l, mid, a, mid, lp), qry(mid + 1, r, mid + 1, b, rp));
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  scanf(" %s", s + 1);
  for (int i = 1; i <= (n); ++i) {
    an[i] = s[i] - '0';
    for (int j = 0; j < (10); ++j) {
      sn[i][j] = sn[i - 1][j];
      if (an[i] == j) ++sn[i][j];
    }
  }
  build(1, n, 1);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    Node ret = qry(1, n, l, r, 1);
    int ans = ret.a[0][4];
    if (ans == 0x3f3f3f3f) ans = -1;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int const N = 202020;
char s[N];
int an[N];
int sn[N][10];
struct Node {
  int a[5][5];
  void init() {
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) a[i][j] = 0x3f3f3f3f;
  }
  void init(int l, int r) {
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) a[i][j] = 0x3f3f3f3f;
    if (sn[r][2] - sn[l - 1][2] > 0) {
      a[0][1] = 0;
      a[0][0] = sn[r][2] - sn[l - 1][2];
    } else {
      a[0][0] = 0;
    }
    if (sn[r][0] - sn[l - 1][0] > 0) {
      a[1][2] = 0;
      a[1][1] = sn[r][0] - sn[l - 1][0];
    } else {
      a[1][1] = 0;
    }
    if (sn[r][1] - sn[l - 1][1] > 0) {
      a[2][3] = 0;
      a[2][2] = sn[r][1] - sn[l - 1][1];
    } else {
      a[2][2] = 0;
    }
    if (sn[r][7] - sn[l - 1][7] > 0) {
      a[3][4] = sn[r][6] - sn[l - 1][6];
      a[3][3] = sn[r][6] - sn[l - 1][6] + sn[r][7] - sn[l - 1][7];
    } else {
      a[3][3] = sn[r][6] - sn[l - 1][6];
    }
    a[4][4] = sn[r][6] - sn[l - 1][6];
  }
  void upd(int l, int r, int c) { a[l][r] = min(a[l][r], c); }
} tree[N << 2];
Node merge(Node A, Node B) {
  Node ret;
  ret.init();
  for (int i = 0; i < (5); ++i)
    for (int j = i; j < 5; ++j) {
      for (int k = i; k <= j; ++k) {
        int lt = A.a[i][k];
        int rt = B.a[k][j];
        ret.upd(i, j, lt + rt);
      }
    }
  return ret;
}
void build(int l, int r, int p) {
  if (l == r) {
    tree[p].init(l, l);
    return;
  }
  int mid = (l + r) >> 1, lp = p << 1, rp = p << 1 | 1;
  build(l, mid, lp);
  build(mid + 1, r, rp);
  tree[p] = merge(tree[lp], tree[rp]);
}
Node qry(int l, int r, int a, int b, int p) {
  if (l == a && r == b) {
    return tree[p];
  }
  int mid = (l + r) >> 1, lp = p << 1, rp = p << 1 | 1;
  if (b <= mid)
    return qry(l, mid, a, b, lp);
  else if (a > mid)
    return qry(mid + 1, r, a, b, rp);
  else {
    return merge(qry(l, mid, a, mid, lp), qry(mid + 1, r, mid + 1, b, rp));
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  scanf(" %s", s + 1);
  for (int i = 1; i <= (n); ++i) {
    an[i] = s[i] - '0';
    for (int j = 0; j < (10); ++j) {
      sn[i][j] = sn[i - 1][j];
      if (an[i] == j) ++sn[i][j];
    }
  }
  build(1, n, 1);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    Node ret = qry(1, n, l, r, 1);
    int ans = ret.a[0][4];
    if (ans == 0x3f3f3f3f) ans = -1;
    printf("%d\n", ans);
  }
  return 0;
}

```