#include <bits/stdc++.h>
using namespace std;
int n;
char s[200010];
int pre6[200010];
struct Inf {
  int trans[5][5];
  void zero() {
    for (int i = 0; i < 5; i++) trans[i][i] = 0;
    for (int i = 0; i < 5; i++)
      for (int j = i + 1; j < 5; j++) trans[i][j] = 1000000;
  }
};
Inf operator+(const Inf& a, const Inf& b) {
  Inf res;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) res.trans[i][j] = 1000000;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++)
      for (int k = j; k < 5; k++) {
        res.trans[i][k] = min(res.trans[i][k], a.trans[i][j] + b.trans[j][k]);
      }
  return res;
}
struct Node {
  int lc, rc;
  Inf a;
} t[400010];
int node;
void pushup(int o) { t[o].a = t[t[o].lc].a + t[t[o].rc].a; }
int build(int l, int r) {
  int now = ++node;
  int mid = (l + r) >> 1;
  if (l == r) {
    t[now].a.zero();
    if (s[l] == '2') {
      t[now].a.trans[0][0] = 1;
      t[now].a.trans[0][1] = 0;
    }
    if (s[l] == '0') {
      t[now].a.trans[1][1] = 1;
      t[now].a.trans[1][2] = 0;
    }
    if (s[l] == '1') {
      t[now].a.trans[2][2] = 1;
      t[now].a.trans[2][3] = 0;
    }
    if (s[l] == '7') {
      t[now].a.trans[3][3] = 1;
      t[now].a.trans[3][4] = 0;
    }
    if (s[l] == '6') {
      t[now].a.trans[3][3] = 1;
      t[now].a.trans[4][4] = 1;
    }
    return now;
  }
  t[now].lc = build(l, mid);
  t[now].rc = build(mid + 1, r);
  pushup(now);
  return now;
}
int query(int l, int r, int ll, int rr, int o) {
  if (l == ll && r == rr) return o;
  int mid = (ll + rr) >> 1;
  if (r <= mid)
    return query(l, r, ll, mid, t[o].lc);
  else if (l > mid)
    return query(l, r, mid + 1, rr, t[o].rc);
  else {
    int now = ++node;
    t[now].lc = query(l, mid, ll, mid, t[o].lc);
    t[now].rc = query(mid + 1, r, mid + 1, rr, t[o].rc);
    pushup(now);
    return now;
  }
}
int savenode;
int query(int l, int r) {
  node = savenode;
  return query(l, r, 1, n, 1);
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  build(1, n);
  savenode = node;
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = query(l, r);
    if (t[ans].a.trans[0][4] > r - l + 1)
      printf("-1\n");
    else
      printf("%d\n", t[ans].a.trans[0][4]);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 n;
char s[200010];
int pre6[200010];
struct Inf {
  int trans[5][5];
  void zero() {
    for (int i = 0; i < 5; i++) trans[i][i] = 0;
    for (int i = 0; i < 5; i++)
      for (int j = i + 1; j < 5; j++) trans[i][j] = 1000000;
  }
};
Inf operator+(const Inf& a, const Inf& b) {
  Inf res;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) res.trans[i][j] = 1000000;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++)
      for (int k = j; k < 5; k++) {
        res.trans[i][k] = min(res.trans[i][k], a.trans[i][j] + b.trans[j][k]);
      }
  return res;
}
struct Node {
  int lc, rc;
  Inf a;
} t[400010];
int node;
void pushup(int o) { t[o].a = t[t[o].lc].a + t[t[o].rc].a; }
int build(int l, int r) {
  int now = ++node;
  int mid = (l + r) >> 1;
  if (l == r) {
    t[now].a.zero();
    if (s[l] == '2') {
      t[now].a.trans[0][0] = 1;
      t[now].a.trans[0][1] = 0;
    }
    if (s[l] == '0') {
      t[now].a.trans[1][1] = 1;
      t[now].a.trans[1][2] = 0;
    }
    if (s[l] == '1') {
      t[now].a.trans[2][2] = 1;
      t[now].a.trans[2][3] = 0;
    }
    if (s[l] == '7') {
      t[now].a.trans[3][3] = 1;
      t[now].a.trans[3][4] = 0;
    }
    if (s[l] == '6') {
      t[now].a.trans[3][3] = 1;
      t[now].a.trans[4][4] = 1;
    }
    return now;
  }
  t[now].lc = build(l, mid);
  t[now].rc = build(mid + 1, r);
  pushup(now);
  return now;
}
int query(int l, int r, int ll, int rr, int o) {
  if (l == ll && r == rr) return o;
  int mid = (ll + rr) >> 1;
  if (r <= mid)
    return query(l, r, ll, mid, t[o].lc);
  else if (l > mid)
    return query(l, r, mid + 1, rr, t[o].rc);
  else {
    int now = ++node;
    t[now].lc = query(l, mid, ll, mid, t[o].lc);
    t[now].rc = query(mid + 1, r, mid + 1, rr, t[o].rc);
    pushup(now);
    return now;
  }
}
int savenode;
int query(int l, int r) {
  node = savenode;
  return query(l, r, 1, n, 1);
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  build(1, n);
  savenode = node;
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = query(l, r);
    if (t[ans].a.trans[0][4] > r - l + 1)
      printf("-1\n");
    else
      printf("%d\n", t[ans].a.trans[0][4]);
  }
  return 0;
}

```