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>
template <class T>
void rd(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
const int N = (1 << 12) + 10;
using namespace std;
int n, q, cnt, a[N], hs[N];
struct Node {
  vector<int> q;
  map<pair<int, int>, int> S;
} f[N << 2];
void build(int o, int l, int r) {
  vector<int> &t = f[o].q;
  if (l == r)
    t.push_back(hs[l]);
  else {
    build((o << 1), l, (l + r >> 1));
    build((o << 1 | 1), (l + r >> 1) + 1, r);
    vector<int> &t1 = f[(o << 1)].q;
    vector<int> &t2 = f[(o << 1 | 1)].q;
    t.resize(r - l + 1);
    merge(t1.begin(), t1.end(), t2.begin(), t2.end(), t.begin());
  }
}
vector<pair<int, int> > op;
vector<int> ans;
int work(int l, int r, int o) {
  vector<int> &t = f[o].q;
  int L = lower_bound(t.begin(), t.end(), l) - t.begin();
  int R = upper_bound(t.begin(), t.end(), r) - t.begin() - 1;
  if (L > R) return -1;
  if (L == R) return t[L];
  map<pair<int, int>, int> &u = f[o].S;
  if (u.find(make_pair(L, R)) == u.end()) {
    int a1 = work(l, r, (o << 1));
    int a2 = work(l, r, (o << 1 | 1));
    if (a1 != -1 && a2 != -1) {
      u[make_pair(L, R)] = ++cnt;
      op.push_back(make_pair(a1, a2));
    } else
      u[make_pair(L, R)] = max(a1, a2);
  }
  return u[make_pair(L, R)];
}
int main() {
  rd(n);
  rd(q);
  cnt = n;
  for (int i = 1; i <= n; i++) {
    rd(a[i]);
    hs[a[i]] = i;
  }
  build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int l, r;
    rd(l);
    rd(r);
    ans.push_back(work(l, r, 1));
  }
  printf("%d\n", cnt);
  for (int i = 0; i < op.size(); i++)
    printf("%d %d\n", op[i].first, op[i].second);
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}