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 <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
using namespace std;
long long ksm(long long a, long long b) {
  if (!b) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 998244353;
  if (b & 1) ns = ns * a % 998244353;
  return ns;
}
int pl[5500];
struct th {
  int sz;
  int l, r;
  vector<int> pls;
  th *ch[2];
  vector<vector<int> > id;
} p[200005], *rt[70];
int cal(th *a, int l, int r) {
  int p1 = lower_bound(a->pls.begin(), a->pls.end(), l) - a->pls.begin();
  int p2 =
      lower_bound(a->pls.begin(), a->pls.end(), r + 1) - a->pls.begin() - 1;
  if (p1 > p2) return 0;
  return a->id[p1][p2];
}
int cnt = 0;
vector<pair<int, int> > g;
int mer(int x, int y) {
  if (!x) return y;
  if (!y) return x;
  cnt++;
  g.push_back(make_pair(x, y));
  return cnt;
}
int stcnt = 0;
void bdtree(th *a, int l, int r) {
  a->l = l, a->r = r;
  for (int i = l; i <= r; i++) a->pls.push_back(pl[i]);
  int sz = r - l + 1;
  sort(a->pls.begin(), a->pls.end());
  a->id.resize(sz);
  for (int j = 0; j < sz; j++) a->id[j].resize(sz);
  if (sz == 1)
    a->id[0][0] = a->pls[0];
  else {
    int mid = (l + r) >> 1;
    a->ch[0] = &p[stcnt++], a->ch[1] = &p[stcnt++];
    bdtree(a->ch[0], l, mid);
    bdtree(a->ch[1], mid + 1, r);
    for (int i = 0; i < sz; i++)
      for (int j = i; j < sz; j++)
        a->id[i][j] = mer(cal(a->ch[0], a->pls[i], a->pls[j]),
                          cal(a->ch[1], a->pls[i], a->pls[j]));
  }
}
int re[100005];
int main() {
  int n, qs;
  scanf("%d%d", &n, &qs);
  cnt = n;
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a), pl[a] = i;
  }
  vector<th *> q;
  int nr = 1;
  while (nr <= n) {
    th *n1 = &p[stcnt++];
    int ur = min(n, nr + 200);
    bdtree(n1, nr, ur);
    q.push_back(n1);
    nr = ur + 1;
  }
  for (int i = 1; i <= qs; i++) {
    int pr = 0;
    int l, r;
    scanf("%d%d", &l, &r);
    for (auto m : q) pr = mer(pr, cal(m, l, r));
    re[i] = pr;
  }
  cout << cnt << endl;
  for (auto h : g) printf("%d %d\n", h.first, h.second);
  for (int i = 1; i <= qs; i++) printf("%d ", re[i]);
  return 0;
}