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 X = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') X = X * 10 + c - '0', c = getchar();
  return X * w;
}
const int N = 4096 + 10;
int n, Q, a[N], pos[N], ans[N << 4];
int B, bl[N], L[N], R[N];
vector<pair<int, int>> vec;
int tot = 0;
int merge(int x, int y) {
  if (!x || !y)
    return x | y;
  else {
    vec.emplace_back(x, y);
    return ++tot;
  }
}
struct Block {
  vector<int> o;
  vector<vector<int>> id;
  int get(int l, int r) {
    int x = lower_bound(o.begin(), o.end(), l) - o.begin();
    int y = upper_bound(o.begin(), o.end(), r) - o.begin() - 1;
    return x <= y ? id[x][y] : 0;
  }
} b[N];
Block merge(Block x, Block y) {
  int n = x.o.size() + y.o.size();
  Block z;
  z.o.resize(n), z.id.resize(n);
  for (int i = 0; i < n; ++i) z.id[i].resize(n);
  merge(x.o.begin(), x.o.end(), y.o.begin(), y.o.end(), z.o.begin());
  for (int i = 0; i < n; ++i)
    for (int j = i; j < n; ++j) {
      int p = x.get(z.o[i], z.o[j]), q = y.get(z.o[i], z.o[j]);
      z.id[i][j] = merge(p, q);
    }
  return z;
}
Block build(int l, int r) {
  if (l == r) return (Block){{pos[l]}, {{pos[l]}}};
  int mid = (l + r) >> 1;
  return merge(build(l, mid), build(mid + 1, r));
}
int main() {
  n = read(), Q = read();
  for (int i = 1; i <= n; ++i) a[i] = read(), pos[a[i]] = i;
  B = sqrt(Q);
  for (int i = 1; i <= n; ++i) bl[i] = (i - 1) / B + 1;
  for (int i = 1; i <= bl[n]; ++i) L[i] = R[i - 1] + 1, R[i] = min(i * B, n);
  tot = n;
  for (int i = 1; i <= bl[n]; ++i) b[i] = build(L[i], R[i]);
  for (int i = 1; i <= Q; ++i) {
    int l = read(), r = read();
    for (int j = 1; j <= bl[n]; ++j) ans[i] = merge(ans[i], b[j].get(l, r));
  }
  printf("%d\n", tot);
  for (auto i : vec) printf("%d %d\n", i.first, i.second);
  for (int i = 1; i <= Q; ++i) printf("%d%c", ans[i], " \n"[i == Q]);
  return 0;
}