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 = 1 << 8, N = (1 << 12) | 5;
int t[B * 2][B | 5][B | 5];
int tt[20][B | 5][B | 5];
int n, Q, a[N], pos[N], ans[N * B];
int ansL[1 << 22], ansR[1 << 22], nd;
int p1[N], p2[N], q[N], p[20][N];
int ADD(int x, int y) {
  if (!x || !y) return x + y;
  ansL[++nd] = x;
  ansR[nd] = y;
  return nd;
}
void output() {
  printf("%d\n", nd);
  for (int i = (int)(n + 1); i <= (int)(nd); i++)
    printf("%d %d\n", ansL[i], ansR[i]);
}
void init(int k, int L, int R) {
  if (L == R) {
    t[k][1][1] = pos[L];
    return;
  }
  int mid = (L + R) / 2;
  init(k * 2, L, mid);
  init(k * 2 + 1, mid + 1, R);
  memset(p1, 0, sizeof(p1));
  memset(p2, 0, sizeof(p2));
  for (int i = (int)(L); i <= (int)(mid); i++) p1[pos[i]]++;
  for (int i = (int)(mid + 1); i <= (int)(R); i++) p2[pos[i]]++;
  for (int i = (int)(1); i <= (int)(n); i++) {
    p1[i] += p1[i - 1];
    p2[i] += p2[i - 1];
  }
  *q = 0;
  for (int i = (int)(L); i <= (int)(R); i++) q[++*q] = pos[i];
  sort(q + 1, q + *q + 1);
  for (int i = (int)(1); i <= (int)(*q); i++)
    for (int j = (int)(i); j <= (int)(*q); j++) {
      int pl1 = p1[q[i] - 1] + 1, pr1 = p1[q[j]];
      int pl2 = p2[q[i] - 1] + 1, pr2 = p2[q[j]];
      int v1 = (pl1 <= pr1 ? t[k * 2][pl1][pr1] : 0);
      int v2 = (pl2 <= pr2 ? t[k * 2 + 1][pl2][pr2] : 0);
      t[k][i][j] = ADD(v1, v2);
    }
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]), pos[a[i]] = i;
  int cnt = (n - 1) / B + 1;
  nd = n;
  for (int i = (int)(1); i <= (int)(cnt); i++) {
    int L = (i - 1) * B + 1, R = min(n, i * B);
    for (int j = (int)(L); j <= (int)(R); j++) p[i][pos[j]] = 1;
    for (int j = (int)(1); j <= (int)(n); j++) p[i][j] += p[i][j - 1];
    init(1, L, R);
    int sum = R - L + 1;
    for (int j = (int)(1); j <= (int)(sum); j++)
      for (int k = (int)(j); k <= (int)(sum); k++) tt[i][j][k] = t[1][j][k];
  }
  for (int i = (int)(1); i <= (int)(Q); i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    ans[i] = 0;
    for (int j = (int)(1); j <= (int)(cnt); j++) {
      int pl = p[j][l - 1] + 1;
      int pr = p[j][r];
      ans[i] = ADD(ans[i], tt[j][pl][pr]);
    }
  }
  output();
  for (int i = (int)(1); i <= (int)(Q); i++) printf("%d ", ans[i]);
}