#include <bits/stdc++.h>
using namespace std;
const int max_n = 222222, inf = 1011111111;
const int arr[] = {2, 0, 1, 6};
int n, q, cnt[max_n][11], pr[max_n][11];
char s[max_n];
int get_cnt(int x, int l, int r) {
  if (l == 0) {
    return cnt[r][x];
  }
  return cnt[r][x] - cnt[l - 1][x];
}
struct state {
  int x[4][4], l, r;
  state() {
    memset(x, 0, sizeof(x));
    l = r = 0;
  }
  state(int pos) {
    memset(x, 0, sizeof(x));
    l = pos;
    r = pos;
    for (int i = 0; i < 4; ++i) {
      if (arr[i] == s[pos]) {
        x[i][i] = 1;
      }
    }
  }
  void write(string s = "") {
    printf("[%d %d] : ", l, r);
    for (int l = 0; l < 4; ++l) {
      for (int r = l; r < 4; ++r) {
        cout << "(" << l << " " << r << " = " << x[l][r] << ") ";
      }
    }
    cout << s;
  }
};
state merge(const state &x, const state &y, int v) {
  state z;
  z.l = x.l;
  z.r = y.r;
  for (int l = 0; l < 4; ++l) {
    z.x[l][l] = x.x[l][l] + y.x[l][l];
    for (int r = l + 1; r < 4; ++r) {
      z.x[l][r] = inf;
      for (int k = r; k >= l; --k) {
        int a = x.x[l][k] + y.x[k][r];
        z.x[l][r] = min(z.x[l][r], a);
      }
    }
  }
  return z;
}
struct tree {
  state a[4 * max_n];
  tree() {}
  void build(int v, int l, int r) {
    if (l == r) {
      a[v] = state(l);
      return;
    }
    int mid = (l + r) / 2;
    build(2 * v, l, mid);
    build(2 * v + 1, mid + 1, r);
    a[v] = merge(a[2 * v], a[2 * v + 1], mid + 1);
  }
  void update(int v, int tl, int tr, int l, int r) {
    if (tl == l && tr == r) {
      return;
    }
    int mid = (tl + tr) / 2;
    if (r <= mid) {
      update(2 * v, tl, mid, l, r);
    } else if (l > mid) {
      update(2 * v + 1, mid + 1, tr, l, r);
    } else {
      update(2 * v, tl, mid, l, mid);
      update(2 * v + 1, mid + 1, tr, mid + 1, r);
    }
    a[v] = merge(a[2 * v], a[2 * v + 1], mid + 1);
  }
  state get(int v, int tl, int tr, int l, int r) {
    if (tl == l && tr == r) {
      return a[v];
    }
    int mid = (tl + tr) / 2;
    if (r <= mid) {
      return get(2 * v, tl, mid, l, r);
    } else if (l > mid) {
      return get(2 * v + 1, mid + 1, tr, l, r);
    }
    return merge(get(2 * v, tl, mid, l, mid),
                 get(2 * v + 1, mid + 1, tr, mid + 1, r), mid + 1);
  }
};
tree t;
int solve(int l, int r) {
  int a, b, c, d;
  d = pr[r][7];
  if (d < l) {
    return -1;
  }
  c = pr[d][1];
  if (c < l) {
    return -1;
  }
  b = pr[c][0];
  if (b < l) {
    return -1;
  }
  a = pr[b][2];
  if (a < l) {
    return -1;
  }
  int res = get_cnt(6, c + 1, r);
  state q = t.get(1, 0, n - 1, l, a);
  state w = t.get(1, 0, n - 1, a, c);
  state Q = t.get(1, 0, n - 1, a, b - 1);
  state W = t.get(1, 0, n - 1, b + 1, c);
  int mn = inf;
  for (int i = 3; i >= 0; --i) {
    int res = q.x[0][i] + w.x[i][3];
    if (i <= 1) {
      res -= w.x[i][3];
      int A = Q.x[1][3] + W.x[3][3];
      int B = Q.x[1][2] + W.x[2][3];
      int C = Q.x[1][1] + W.x[2][3];
      res += min({A, B, C});
    }
    mn = min(mn, res);
  }
  return res + mn;
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", s);
  memset(pr, -1, sizeof(pr));
  for (int i = 0; i < n; ++i) {
    s[i] -= '0';
    if (i) {
      for (int j = 0; j < 10; ++j) {
        cnt[i][j] = cnt[i - 1][j];
        pr[i][j] = pr[i - 1][j];
      }
    }
    ++cnt[i][s[i]];
    pr[i][s[i]] = i;
  }
  t.build(1, 0, n - 1);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l;
    --r;
    printf("%d\n", solve(l, r));
  }
  return 0;
}

### Prompt

Please formulate 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 max_n = 222222, inf = 1011111111;
const int arr[] = {2, 0, 1, 6};
int n, q, cnt[max_n][11], pr[max_n][11];
char s[max_n];
int get_cnt(int x, int l, int r) {
  if (l == 0) {
    return cnt[r][x];
  }
  return cnt[r][x] - cnt[l - 1][x];
}
struct state {
  int x[4][4], l, r;
  state() {
    memset(x, 0, sizeof(x));
    l = r = 0;
  }
  state(int pos) {
    memset(x, 0, sizeof(x));
    l = pos;
    r = pos;
    for (int i = 0; i < 4; ++i) {
      if (arr[i] == s[pos]) {
        x[i][i] = 1;
      }
    }
  }
  void write(string s = "") {
    printf("[%d %d] : ", l, r);
    for (int l = 0; l < 4; ++l) {
      for (int r = l; r < 4; ++r) {
        cout << "(" << l << " " << r << " = " << x[l][r] << ") ";
      }
    }
    cout << s;
  }
};
state merge(const state &x, const state &y, int v) {
  state z;
  z.l = x.l;
  z.r = y.r;
  for (int l = 0; l < 4; ++l) {
    z.x[l][l] = x.x[l][l] + y.x[l][l];
    for (int r = l + 1; r < 4; ++r) {
      z.x[l][r] = inf;
      for (int k = r; k >= l; --k) {
        int a = x.x[l][k] + y.x[k][r];
        z.x[l][r] = min(z.x[l][r], a);
      }
    }
  }
  return z;
}
struct tree {
  state a[4 * max_n];
  tree() {}
  void build(int v, int l, int r) {
    if (l == r) {
      a[v] = state(l);
      return;
    }
    int mid = (l + r) / 2;
    build(2 * v, l, mid);
    build(2 * v + 1, mid + 1, r);
    a[v] = merge(a[2 * v], a[2 * v + 1], mid + 1);
  }
  void update(int v, int tl, int tr, int l, int r) {
    if (tl == l && tr == r) {
      return;
    }
    int mid = (tl + tr) / 2;
    if (r <= mid) {
      update(2 * v, tl, mid, l, r);
    } else if (l > mid) {
      update(2 * v + 1, mid + 1, tr, l, r);
    } else {
      update(2 * v, tl, mid, l, mid);
      update(2 * v + 1, mid + 1, tr, mid + 1, r);
    }
    a[v] = merge(a[2 * v], a[2 * v + 1], mid + 1);
  }
  state get(int v, int tl, int tr, int l, int r) {
    if (tl == l && tr == r) {
      return a[v];
    }
    int mid = (tl + tr) / 2;
    if (r <= mid) {
      return get(2 * v, tl, mid, l, r);
    } else if (l > mid) {
      return get(2 * v + 1, mid + 1, tr, l, r);
    }
    return merge(get(2 * v, tl, mid, l, mid),
                 get(2 * v + 1, mid + 1, tr, mid + 1, r), mid + 1);
  }
};
tree t;
int solve(int l, int r) {
  int a, b, c, d;
  d = pr[r][7];
  if (d < l) {
    return -1;
  }
  c = pr[d][1];
  if (c < l) {
    return -1;
  }
  b = pr[c][0];
  if (b < l) {
    return -1;
  }
  a = pr[b][2];
  if (a < l) {
    return -1;
  }
  int res = get_cnt(6, c + 1, r);
  state q = t.get(1, 0, n - 1, l, a);
  state w = t.get(1, 0, n - 1, a, c);
  state Q = t.get(1, 0, n - 1, a, b - 1);
  state W = t.get(1, 0, n - 1, b + 1, c);
  int mn = inf;
  for (int i = 3; i >= 0; --i) {
    int res = q.x[0][i] + w.x[i][3];
    if (i <= 1) {
      res -= w.x[i][3];
      int A = Q.x[1][3] + W.x[3][3];
      int B = Q.x[1][2] + W.x[2][3];
      int C = Q.x[1][1] + W.x[2][3];
      res += min({A, B, C});
    }
    mn = min(mn, res);
  }
  return res + mn;
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", s);
  memset(pr, -1, sizeof(pr));
  for (int i = 0; i < n; ++i) {
    s[i] -= '0';
    if (i) {
      for (int j = 0; j < 10; ++j) {
        cnt[i][j] = cnt[i - 1][j];
        pr[i][j] = pr[i - 1][j];
      }
    }
    ++cnt[i][s[i]];
    pr[i][s[i]] = i;
  }
  t.build(1, 0, n - 1);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l;
    --r;
    printf("%d\n", solve(l, r));
  }
  return 0;
}

```