#include <bits/stdc++.h>
#pragma GCC optimize(3, "Ofast", "inline")
#pragma GCC target("avx,avx2")
using namespace std;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 2e5 + 5;
int a[N], n, m;
struct node {
  int f[5][5];
  void modify(int x) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) f[i][j] = i == j ? 0 : 0x3f3f3f3f;
    if (x == 2) f[0][0] = 1, f[0][1] = 0;
    if (x == 0) f[1][1] = 1, f[1][2] = 0;
    if (x == 1) f[2][2] = 1, f[2][3] = 0;
    if (x == 7) f[3][3] = 1, f[3][4] = 0;
    if (x == 6) f[3][3] = 1, f[4][4] = 1;
  }
  inline node operator+(const node &nt) const {
    node res;
    memset(res.f, 0x3f, sizeof res.f);
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++)
        for (int k = 0; k < 5; k++)
          res.f[i][j] = min(res.f[i][j], f[i][k] + nt.f[k][j]);
    return res;
  }
} tr[N << 2];
void build(int x, int l, int r) {
  if (l == r) {
    tr[x].modify(a[l]);
    return;
  }
  int mid = l + r >> 1;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
  tr[x] = tr[x << 1] + tr[x << 1 | 1];
}
node que(int x, int l, int r, int p, int q) {
  if (p <= l && r <= q) return tr[x];
  int op = 0, mid = l + r >> 1;
  node A, B;
  if (p <= mid) op++, A = que(x << 1, l, mid, p, q);
  if (q > mid) op += 2, B = que(x << 1 | 1, mid + 1, r, p, q);
  if (op == 1) return A;
  if (op == 2) return B;
  return A + B;
}
signed main() {
  read(n, m);
  for (int i = 1; i <= n; i++) scanf("%1d", &a[i]);
  build(1, 1, n);
  while (m--) {
    int l, r;
    read(l, r);
    int ans = que(1, 1, n, l, r).f[0][4];
    if (ans == 0x3f3f3f3f) ans = -1;
    write(ans);
    puts("");
  }
}

### 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>
#pragma GCC optimize(3, "Ofast", "inline")
#pragma GCC target("avx,avx2")
using namespace std;
template <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const int N = 2e5 + 5;
int a[N], n, m;
struct node {
  int f[5][5];
  void modify(int x) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) f[i][j] = i == j ? 0 : 0x3f3f3f3f;
    if (x == 2) f[0][0] = 1, f[0][1] = 0;
    if (x == 0) f[1][1] = 1, f[1][2] = 0;
    if (x == 1) f[2][2] = 1, f[2][3] = 0;
    if (x == 7) f[3][3] = 1, f[3][4] = 0;
    if (x == 6) f[3][3] = 1, f[4][4] = 1;
  }
  inline node operator+(const node &nt) const {
    node res;
    memset(res.f, 0x3f, sizeof res.f);
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++)
        for (int k = 0; k < 5; k++)
          res.f[i][j] = min(res.f[i][j], f[i][k] + nt.f[k][j]);
    return res;
  }
} tr[N << 2];
void build(int x, int l, int r) {
  if (l == r) {
    tr[x].modify(a[l]);
    return;
  }
  int mid = l + r >> 1;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
  tr[x] = tr[x << 1] + tr[x << 1 | 1];
}
node que(int x, int l, int r, int p, int q) {
  if (p <= l && r <= q) return tr[x];
  int op = 0, mid = l + r >> 1;
  node A, B;
  if (p <= mid) op++, A = que(x << 1, l, mid, p, q);
  if (q > mid) op += 2, B = que(x << 1 | 1, mid + 1, r, p, q);
  if (op == 1) return A;
  if (op == 2) return B;
  return A + B;
}
signed main() {
  read(n, m);
  for (int i = 1; i <= n; i++) scanf("%1d", &a[i]);
  build(1, 1, n);
  while (m--) {
    int l, r;
    read(l, r);
    int ans = que(1, 1, n, l, r).f[0][4];
    if (ans == 0x3f3f3f3f) ans = -1;
    write(ans);
    puts("");
  }
}

```