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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, Q, len, blocks, L[75010], R[75010], a[75010], pos[75010], cnt;
vector<pair<int, int> > OP;
vector<int> p[75010], ans;
vector<vector<int> > q[75010];
int Merge(int x, int y) {
  if (!x || !y) return x | y;
  ++cnt;
  OP.emplace_back(x, y);
  return cnt;
}
int Query(int l, int r, const vector<int> &V, const vector<vector<int> > &T) {
  l = lower_bound(V.begin(), V.end(), l) - V.begin();
  r = upper_bound(V.begin(), V.end(), r) - V.begin() - 1;
  if (l >= V.size() || r < 0 || l > r) return 0;
  return T[l][r];
}
vector<vector<int> > Solve(vector<int> &V) {
  vector<vector<int> > T;
  T.resize(V.size());
  for (int i = 0; i < (int)V.size(); ++i) {
    T[i].resize(V.size());
  }
  if (V.size() == 1) {
    T[0][0] = V[0];
    return T;
  }
  auto tmp = V;
  nth_element(tmp.begin(), tmp.begin() + (int)V.size() / 2, tmp.end(),
              [&](int i, int j) { return a[i] < a[j]; });
  int mid = tmp[V.size() / 2];
  vector<int> A, B;
  for (int i = 0; i < (int)V.size(); ++i) {
    if (a[V[i]] < a[mid])
      A.push_back(V[i]);
    else
      B.push_back(V[i]);
  }
  auto TA = Solve(A), TB = Solve(B);
  for (int i = 0; i < (int)V.size(); ++i) {
    for (int j = i; j < (int)V.size(); ++j) {
      T[i][j] = Merge(Query(V[i], V[j], A, TA), Query(V[i], V[j], B, TB));
    }
  }
  return T;
}
int main() {
  n = read(), Q = read();
  cnt = n;
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
  }
  len = min(n, (int)ceil(sqrt(Q)));
  blocks = n / len;
  if (n % len) ++blocks;
  for (int i = 1; i <= blocks; ++i) {
    L[i] = (i - 1) * len + 1, R[i] = min(n, i * len);
    for (int j = L[i]; j <= R[i]; ++j) pos[j] = i;
  }
  for (int i = 1; i <= blocks; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (a[j] >= L[i] && a[j] <= R[i]) p[i].push_back(j);
    }
    q[i] = Solve(p[i]);
  }
  while (Q--) {
    int l = read(), r = read();
    int now = 0;
    for (int i = 1; i <= blocks; ++i) {
      now = Merge(now, Query(l, r, p[i], q[i]));
    }
    ans.push_back(now);
  }
  printf("%d\n", n + (int)OP.size());
  for (auto [x, y] : OP) printf("%d %d\n", x, y);
  for (auto x : ans) printf("%d ", x);
  printf("\n");
  return 0;
}