Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int B = 256;
int f[B << 1 | 1][B | 3][B | 3], cnt, n, m, ans[70005], b, a[5005], pos[5005],
    stp[2200005][2];
int insert(int x, int y) {
  if (!x || !y) return x + y;
  return stp[++cnt][0] = x, stp[cnt][1] = y, cnt;
}
struct Block {
  int anss[B | 2][B | 2], s[5005];
  void build(int k, int l, int r, vector<int> vec) {
    if (l == r) return void(f[k][1][1] = pos[l]);
    vector<int> vecl(1), vecr(1);
    int mid = l + r >> 1;
    for (int i = 1; i < vec.size(); i++)
      (vec[i] <= mid) ? (vecl.push_back(vec[i])) : (vecr.push_back(vec[i]));
    build(k << 1, l, mid, vecl), build(k << 1 | 1, mid + 1, r, vecr);
    for (int i = 1, xl = 1, xr = 1; i < vec.size();
         (vec[i] <= mid) ? (xl++) : (xr++), i++)
      for (int j = i, yl = xl - (vec[j] > mid), yr = xr - (vec[j] <= mid);
           j < vec.size(); j++, (vec[j] <= mid) ? (yl++) : (yr++))
        f[k][i][j] = insert(f[k << 1][xl][yl], f[k << 1 | 1][xr][yr]);
    return;
  }
  void build1(int l, int r) {
    vector<int> vec(1);
    for (int i = 1; i <= n; i++, s[i] = s[i - 1])
      if (a[i] >= l && a[i] <= r) vec.push_back(a[i]), ++s[i];
    build(1, l, r, vec);
    for (int i = 1; i < vec.size(); i++)
      for (int j = i; j < vec.size(); j++) anss[i][j] = f[1][i][j];
  }
  int query(int l, int r) { return anss[s[l - 1] + 1][s[r]]; }
} block[20];
int main() {
  scanf("%d%d", &n, &m), b = min(B, n), cnt = n;
  for (int i = 1; i <= n; i++) scanf("%d", a + i), pos[a[i]] = i;
  for (int i = 1; (i - 1) * b + 1 <= n; i++)
    block[i].build1((i - 1) * b + 1, min(i * b, n));
  for (int i = 1, l, r; i <= m; i++) {
    scanf("%d%d", &l, &r);
    for (int j = 1; (j - 1) * b + 1 <= n; j++)
      ans[i] = insert(ans[i], block[j].query(l, r));
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; i++) printf("%d %d\n", stp[i][0], stp[i][1]);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}