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 i64 = int64_t;
using u64 = uint64_t;
using u32 = uint32_t;
int n, q;
constexpr int N = 5000;
int a[N], p[N], f[131072];
std::vector<std::pair<int, int>> ans;
struct SegmentTree {
  std::vector<int> pos[N * 4];
  std::map<std::pair<int, int>, int> map[N * 4];
  void build(int l, int r, int p) {
    if (l == r) {
      return void(pos[p].emplace_back(::p[l]));
    } else {
      int mid = l + r >> 1;
      int ls = p << 1;
      int rs = p << 1 | 1;
      build(l, mid, ls);
      build(mid + 1, r, rs);
      int p0(0), p1(0), p2(0);
      pos[p].resize(r - l + 1);
      while (p0 < pos[ls].size() && p1 < pos[rs].size()) {
        pos[p][p2++] =
            pos[ls][p0] < pos[rs][p1] ? pos[ls][p0++] : pos[rs][p1++];
      }
      while (p0 < pos[ls].size()) {
        pos[p][p2++] = pos[ls][p0++];
      }
      while (p1 < pos[rs].size()) {
        pos[p][p2++] = pos[rs][p1++];
      }
    }
  }
  int solve(int L, int R, int p) {
    int l = std::lower_bound(pos[p].begin(), pos[p].end(), L) - pos[p].begin();
    int r =
        std::upper_bound(pos[p].begin(), pos[p].end(), R) - pos[p].begin() - 1;
    if (l > r) {
      return 0;
    }
    if (l == r) {
      return pos[p][l];
    } else if (map[p].count({l, r})) {
      return map[p][{l, r}];
    } else {
      int x = solve(L, R, p << 1);
      int y = solve(L, R, p << 1 | 1);
      if (!x || !y) {
        return map[p][{l, r}] = x | y;
      } else {
        ans.emplace_back(x, y);
        return map[p][{l, r}] = (int)ans.size() + n;
      }
    }
  }
} T;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    std::cin >> a[i];
    p[a[i]] = i;
  }
  T.build(1, n, 1);
  for (int i = 1; i <= q; i++) {
    int l, r;
    std::cin >> l >> r;
    f[i] = T.solve(l, r, 1);
  }
  std::cout << (int)ans.size() + n << "\n";
  for (auto i : ans) {
    std::cout << i.first << " " << i.second << "\n";
  }
  for (int i = 1; i <= q; i++) {
    std::cout << f[i] << " ";
  }
  return 0;
}