#include <bits/stdc++.h>
const int MOD = 1000000007;
const int INF = 21000001;
const double EPS = 1e-6;
const int HASH_POW = 29;
const double PI = acos(-1.0);
using namespace std;
int n, q;
struct node {
  int p[5][5];
} t[800010];
char s[200010];
void merge(node &a, node &b, node &c) {
  for (int i = 0; i < 5; ++i)
    for (int j = 0; j < 5; ++j) c.p[i][j] = INF;
  for (int i = 0; i < 5; ++i) {
    for (int j = i; j < 5; ++j) {
      for (int k = i; k <= j; ++k)
        c.p[i][j] = min(c.p[i][j], a.p[i][k] + b.p[k][j]);
    }
  }
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) t[v].p[i][j] = INF;
    for (int i = 0; i < 5; ++i) t[v].p[i][i] = 0;
    if (s[tl] == '2') {
      t[v].p[0][0] = 1;
      t[v].p[0][1] = 0;
    } else if (s[tl] == '0') {
      t[v].p[1][1] = 1;
      t[v].p[1][2] = 0;
    } else if (s[tl] == '1') {
      t[v].p[2][2] = 1;
      t[v].p[2][3] = 0;
    } else if (s[tl] == '6') {
      t[v].p[3][3] = 1;
      t[v].p[4][4] = 1;
    } else if (s[tl] == '7') {
      t[v].p[3][4] = 0;
    }
  } else {
    int mid = ((tl + tr) >> 1);
    build((v << 1), tl, mid);
    build((v << 1) + 1, mid + 1, tr);
    merge(t[(v << 1)], t[(v << 1) + 1], t[v]);
  }
}
node query(int v, int tl, int tr, int l, int r) {
  if (l > r || tl > r || tr < l) {
    node res;
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) res.p[i][j] = 0;
    return res;
  }
  int mid = ((tl + tr) >> 1);
  node a, b, ret;
  if (tl >= l && tr <= r)
    ret = t[v];
  else {
    a = query((v << 1), tl, mid, l, r);
    b = query((v << 1) + 1, mid + 1, tr, l, r);
    merge(a, b, ret);
    if (r <= mid) return a;
    if (l > mid) return b;
  }
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
    }
  return ret;
}
int main() {
  scanf("%d %d\n%s", &n, &q, s + 1);
  build(1, 1, n);
  while (q-- > 0) {
    int l, r;
    scanf("%d %d", &l, &r);
    int res = query(1, 1, n, l, r).p[0][4];
    if (res == INF)
      printf("-1\n");
    else
      printf("%d\n", res);
  }
  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>
const int MOD = 1000000007;
const int INF = 21000001;
const double EPS = 1e-6;
const int HASH_POW = 29;
const double PI = acos(-1.0);
using namespace std;
int n, q;
struct node {
  int p[5][5];
} t[800010];
char s[200010];
void merge(node &a, node &b, node &c) {
  for (int i = 0; i < 5; ++i)
    for (int j = 0; j < 5; ++j) c.p[i][j] = INF;
  for (int i = 0; i < 5; ++i) {
    for (int j = i; j < 5; ++j) {
      for (int k = i; k <= j; ++k)
        c.p[i][j] = min(c.p[i][j], a.p[i][k] + b.p[k][j]);
    }
  }
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) t[v].p[i][j] = INF;
    for (int i = 0; i < 5; ++i) t[v].p[i][i] = 0;
    if (s[tl] == '2') {
      t[v].p[0][0] = 1;
      t[v].p[0][1] = 0;
    } else if (s[tl] == '0') {
      t[v].p[1][1] = 1;
      t[v].p[1][2] = 0;
    } else if (s[tl] == '1') {
      t[v].p[2][2] = 1;
      t[v].p[2][3] = 0;
    } else if (s[tl] == '6') {
      t[v].p[3][3] = 1;
      t[v].p[4][4] = 1;
    } else if (s[tl] == '7') {
      t[v].p[3][4] = 0;
    }
  } else {
    int mid = ((tl + tr) >> 1);
    build((v << 1), tl, mid);
    build((v << 1) + 1, mid + 1, tr);
    merge(t[(v << 1)], t[(v << 1) + 1], t[v]);
  }
}
node query(int v, int tl, int tr, int l, int r) {
  if (l > r || tl > r || tr < l) {
    node res;
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) res.p[i][j] = 0;
    return res;
  }
  int mid = ((tl + tr) >> 1);
  node a, b, ret;
  if (tl >= l && tr <= r)
    ret = t[v];
  else {
    a = query((v << 1), tl, mid, l, r);
    b = query((v << 1) + 1, mid + 1, tr, l, r);
    merge(a, b, ret);
    if (r <= mid) return a;
    if (l > mid) return b;
  }
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) {
    }
  return ret;
}
int main() {
  scanf("%d %d\n%s", &n, &q, s + 1);
  build(1, 1, n);
  while (q-- > 0) {
    int l, r;
    scanf("%d %d", &l, &r);
    int res = query(1, 1, n, l, r).p[0][4];
    if (res == INF)
      printf("-1\n");
    else
      printf("%d\n", res);
  }
  return 0;
}

```