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;
int read() {
  int f = 0, x = 0;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while ('0' <= c && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return f ? -x : x;
}
const int MAXN = 4096, MAXQ = 65536, B = 326;
vector<short> Val[8 * MAXN + 5];
vector<vector<int> > Id[8 * MAXN + 5];
int a[MAXN + 5], pos[MAXN + 5];
int ncnt, tot, c[(1 << 23) + 5], d[(1 << 23) + 5];
int Merge(int u, int v) {
  if (!u || !v) return u + v;
  c[++tot] = u, d[tot] = v;
  return tot;
}
int rt[B + 5], ch[5 * MAXN + 5][2];
void Init(int u, int len) {
  Val[u].resize(len), Id[u].resize(len);
  for (int i = 0; i < len; i++) Id[u][i].resize(len - i);
  return;
}
int Ask(int u, int L, int R) {
  if (R < Val[u].front() || Val[u].back() < L) return 0;
  L = lower_bound(Val[u].begin(), Val[u].end(), L) - Val[u].begin();
  R = upper_bound(Val[u].begin(), Val[u].end(), R) - Val[u].begin() - 1;
  return L > R ? 0 : Id[u][L][R - L];
}
void Build(int &u, int L, int R) {
  if (!u) u = ++ncnt;
  if (L == R) {
    Init(u, 1);
    Val[u][0] = pos[L];
    Id[u][0][0] = pos[L];
    return;
  }
  int Mid = (L + R) >> 1;
  Build(ch[u][0], L, Mid), Build(ch[u][1], Mid + 1, R);
  Init(u, Val[ch[u][0]].size() + Val[ch[u][1]].size());
  merge(Val[ch[u][0]].begin(), Val[ch[u][0]].end(), Val[ch[u][1]].begin(),
        Val[ch[u][1]].end(), Val[u].begin());
  int siz = Val[u].size();
  for (int i = 0; i < siz; i++)
    for (int j = i; j < siz; j++) {
      int t1 = Ask(ch[u][0], Val[u][i], Val[u][j]);
      int t2 = Ask(ch[u][1], Val[u][i], Val[u][j]);
      Id[u][i][j - i] = Merge(t1, t2);
    }
  return;
}
int Ans[MAXQ + 5];
int main() {
  int n = read(), q = read();
  tot = n;
  for (int i = 1; i <= n; i++) pos[a[i] = read()] = i;
  for (int i = 0; i * B <= n; i++)
    Build(rt[i + 1], i * B + 1, min(n, (i + 1) * B));
  for (int i = 1; i <= q; i++) {
    int L = read(), R = read();
    for (int j = 0; j * B <= n; j++)
      Ans[i] = Merge(Ans[i], Ask(rt[j + 1], L, R));
  }
  printf("%d\n", tot);
  for (int i = n + 1; i <= tot; i++) printf("%d %d\n", c[i], d[i]);
  for (int i = 1; i <= q; i++)
    printf("%d", Ans[i]), putchar(i == q ? '\n' : ' ');
  return 0;
}