#include <bits/stdc++.h>
using namespace std;
namespace Solve {
const int MAXN = 2e5 + 10, INF = MAXN + 1;
int n, m;
char str[MAXN];
struct Info {
  int trans[5][5];
};
Info tree[MAXN << 2];
int debug = 0;
Info merge(const Info& A, const Info& B) {
  Info ans;
  for (int x1 = 0; x1 < 5; x1++) {
    for (int x2 = 0; x2 < 5; x2++) {
      ans.trans[x1][x2] = INF;
    }
  }
  for (int x1 = 0; x1 < 5; x1++) {
    for (int x2 = x1; x2 < 5; x2++) {
      ans.trans[x1][x2] = INF;
      for (int mx = x1; mx <= x2; mx++) {
        ans.trans[x1][x2] =
            min(ans.trans[x1][x2], A.trans[x1][mx] + B.trans[mx][x2]);
      }
    }
  }
  if (debug) {
    for (int mx = 0; mx < 5; mx++) {
      if (A.trans[0][mx] + B.trans[mx][4] == ans.trans[0][4]) {
        cout << "debug : " << mx << " " << A.trans[0][mx] << " "
             << B.trans[mx][4] << endl;
        return ans;
      }
    }
  }
  return ans;
}
void build(int root, int l, int r) {
  if (l == r) {
    for (int x1 = 0; x1 < 5; x1++) {
      int nx = x1;
      if (x1 == 0 && str[l] == '2') nx++;
      if (x1 == 1 && str[l] == '0') nx++;
      if (x1 == 2 && str[l] == '1') nx++;
      if (x1 == 3 && str[l] == '7') nx++;
      if (str[l] == '6' && x1 == 3) {
        for (int x2 = 0; x2 < 5; x2++) {
          tree[root].trans[x1][x2] = INF;
        }
        tree[root].trans[x1][x1] = 1;
      } else if (str[l] == '6' && x1 == 4) {
        for (int x2 = 0; x2 < 5; x2++) {
          tree[root].trans[x1][x2] = INF;
        }
        tree[root].trans[x1][x1] = 1;
      } else {
        for (int x2 = 0; x2 < 5; x2++) {
          tree[root].trans[x1][x2] = INF;
          if (nx == x2) tree[root].trans[x1][nx] = 0;
        }
      }
      tree[root].trans[x1][x1] = min(tree[root].trans[x1][x1], 1);
    }
    return;
  }
  int mid = (l + r) / 2;
  build(root << 1, l, mid);
  build(root << 1 | 1, mid + 1, r);
  tree[root] = merge(tree[root << 1], tree[root << 1 | 1]);
}
Info query(int L, int R, int root, int l, int r) {
  if (L <= l && r <= R) return tree[root];
  int mid = (l + r) / 2;
  if (R <= mid) return query(L, R, root << 1, l, mid);
  if (L > mid) return query(L, R, root << 1 | 1, mid + 1, r);
  return merge(query(L, R, root << 1, l, mid),
               query(L, R, root << 1 | 1, mid + 1, r));
}
void solve() {
  scanf("%d%d", &n, &m);
  scanf("%s", str + 1);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int L, R;
    scanf("%d%d", &L, &R);
    Info ans = query(L, R, 1, 1, n);
    printf("%d\n", ans.trans[0][4] < INF ? ans.trans[0][4] : -1);
  }
}
}  // namespace Solve
int main() {
  Solve::solve();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
namespace Solve {
const int MAXN = 2e5 + 10, INF = MAXN + 1;
int n, m;
char str[MAXN];
struct Info {
  int trans[5][5];
};
Info tree[MAXN << 2];
int debug = 0;
Info merge(const Info& A, const Info& B) {
  Info ans;
  for (int x1 = 0; x1 < 5; x1++) {
    for (int x2 = 0; x2 < 5; x2++) {
      ans.trans[x1][x2] = INF;
    }
  }
  for (int x1 = 0; x1 < 5; x1++) {
    for (int x2 = x1; x2 < 5; x2++) {
      ans.trans[x1][x2] = INF;
      for (int mx = x1; mx <= x2; mx++) {
        ans.trans[x1][x2] =
            min(ans.trans[x1][x2], A.trans[x1][mx] + B.trans[mx][x2]);
      }
    }
  }
  if (debug) {
    for (int mx = 0; mx < 5; mx++) {
      if (A.trans[0][mx] + B.trans[mx][4] == ans.trans[0][4]) {
        cout << "debug : " << mx << " " << A.trans[0][mx] << " "
             << B.trans[mx][4] << endl;
        return ans;
      }
    }
  }
  return ans;
}
void build(int root, int l, int r) {
  if (l == r) {
    for (int x1 = 0; x1 < 5; x1++) {
      int nx = x1;
      if (x1 == 0 && str[l] == '2') nx++;
      if (x1 == 1 && str[l] == '0') nx++;
      if (x1 == 2 && str[l] == '1') nx++;
      if (x1 == 3 && str[l] == '7') nx++;
      if (str[l] == '6' && x1 == 3) {
        for (int x2 = 0; x2 < 5; x2++) {
          tree[root].trans[x1][x2] = INF;
        }
        tree[root].trans[x1][x1] = 1;
      } else if (str[l] == '6' && x1 == 4) {
        for (int x2 = 0; x2 < 5; x2++) {
          tree[root].trans[x1][x2] = INF;
        }
        tree[root].trans[x1][x1] = 1;
      } else {
        for (int x2 = 0; x2 < 5; x2++) {
          tree[root].trans[x1][x2] = INF;
          if (nx == x2) tree[root].trans[x1][nx] = 0;
        }
      }
      tree[root].trans[x1][x1] = min(tree[root].trans[x1][x1], 1);
    }
    return;
  }
  int mid = (l + r) / 2;
  build(root << 1, l, mid);
  build(root << 1 | 1, mid + 1, r);
  tree[root] = merge(tree[root << 1], tree[root << 1 | 1]);
}
Info query(int L, int R, int root, int l, int r) {
  if (L <= l && r <= R) return tree[root];
  int mid = (l + r) / 2;
  if (R <= mid) return query(L, R, root << 1, l, mid);
  if (L > mid) return query(L, R, root << 1 | 1, mid + 1, r);
  return merge(query(L, R, root << 1, l, mid),
               query(L, R, root << 1 | 1, mid + 1, r));
}
void solve() {
  scanf("%d%d", &n, &m);
  scanf("%s", str + 1);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int L, R;
    scanf("%d%d", &L, &R);
    Info ans = query(L, R, 1, 1, n);
    printf("%d\n", ans.trans[0][4] < INF ? ans.trans[0][4] : -1);
  }
}
}  // namespace Solve
int main() {
  Solve::solve();
  return 0;
}

```