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>
const int N = 5e3, Q = 7e4;
int n, q, a[N], ans[Q];
std::vector<std::pair<int, int>> move;
struct data {
  std::vector<int> v;
  std::map<std::pair<int, int>, int> map;
} node[N * 4];
void build(int l, int r, int id = 1) {
  std::vector<int> &v = node[id].v;
  if (l == r) {
    v.push_back(a[l]);
    return;
  }
  int m = (l + r) / 2;
  build(l, m, id * 2);
  build(m + 1, r, id * 2 + 1);
  std::vector<int> &v1 = node[id * 2].v;
  std::vector<int> &v2 = node[id * 2 + 1].v;
  v.resize(r - l + 1);
  std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(), v.begin());
}
int solve(int l, int r, int id = 1) {
  std::vector<int> &v = node[id].v;
  int x = std::lower_bound(v.begin(), v.end(), l) - v.begin();
  int y = std::upper_bound(v.begin(), v.end(), r) - v.begin();
  if (x == v.size() || x == y) return -1;
  if (x + 1 == y) return v[x];
  std::map<std::pair<int, int>, int> &map = node[id].map;
  if (map.find({x, y}) == map.end()) {
    int a = solve(l, r, id * 2);
    int b = solve(l, r, id * 2 + 1);
    if (a != -1 && b != -1) {
      move.push_back({a, b});
      map[{x, y}] = move.size() + n;
    } else
      map[{x, y}] = std::max(a, b);
  }
  return map[{x, y}];
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    int val;
    scanf("%d", &val);
    a[val] = i;
  }
  build(1, n);
  for (int i = 1; i <= q; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    ans[i] = solve(l, r);
  }
  printf("%d\n", move.size() + n);
  for (std::pair<int, int> p : move) printf("%d %d\n", p.first, p.second);
  for (int i = 1; i <= q; ++i) printf("%d ", ans[i]);
  printf("\n");
}