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;
long long read() {
  long long f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -f;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + (s - '0');
    s = getchar();
  }
  return x * f;
}
const int MOD = 1000000007;
int n, m, i, j, s, o, k;
int sq, cnt, recx[2200005], recy[2200005];
int merg(int x, int y) {
  if (!x || !y) return x + y;
  ++cnt;
  recx[cnt] = x;
  recy[cnt] = y;
  return cnt;
}
struct it {
  int nm, id;
  it() { nm = id = 0; }
  it(int N, int I) {
    nm = N;
    id = I;
  }
} a[4105];
bool cmp(it a, it b) { return a.nm < b.nm; }
bool operator<(it a, it b) { return a.id < b.id; }
bool operator>(it a, it b) { return b < a; }
struct blo {
  int l, r;
  vector<int> b;
  map<int, map<int, int> > mp;
  blo() {
    l = r = 0;
    b.clear();
    mp.clear();
  }
} B[4105], tre[4105 << 2];
void makeblock(int A, int l, int r) {
  tre[A] = blo();
  if (l == r) {
    tre[A].b.push_back(a[l].id);
    tre[A].mp[0][0] = a[l].id;
    tre[A].l = tre[A].r = l;
    return;
  }
  int mid = (l + r) >> 1, x, y, a = A;
  makeblock(x = a << 1, l, mid);
  makeblock(y = a << 1 | 1, mid + 1, r);
  tre[a].l = l, tre[a].r = r;
  int i = 0, j = 0;
  while (i < (int)tre[x].b.size() || j < (int)tre[y].b.size()) {
    if (j >= (int)tre[y].b.size() ||
        (i < (int)tre[x].b.size() && tre[x].b[i] <= tre[y].b[j])) {
      if (i >= tre[x].b.size()) printf("Shit !!"), exit(0);
      tre[a].b.push_back(tre[x].b[i++]);
    } else {
      if (j >= tre[y].b.size()) printf("SHIT !!"), exit(0);
      tre[a].b.push_back(tre[y].b[j++]);
    }
  }
  int le = (int)tre[a].b.size();
  for (int i = 0; i < le; i++) {
    for (int j = i; j < le; j++) {
      int lp = tre[a].b[i], rp = tre[a].b[j] + 1;
      int l1 =
          lower_bound(tre[x].b.begin(), tre[x].b.end(), lp) - tre[x].b.begin();
      int r1 = lower_bound(tre[x].b.begin(), tre[x].b.end(), rp) -
               tre[x].b.begin() - 1;
      int l2 =
          lower_bound(tre[y].b.begin(), tre[y].b.end(), lp) - tre[y].b.begin();
      int r2 = lower_bound(tre[y].b.begin(), tre[y].b.end(), rp) -
               tre[y].b.begin() - 1;
      tre[a].mp[i][j] = merg(tre[x].mp[l1][r1], tre[y].mp[l2][r2]);
    }
  }
  tre[x] = tre[y] = blo();
  return;
}
int S[65545];
int main() {
  n = read();
  m = read();
  cnt = n;
  sq = (int)sqrt((double)m);
  for (int i = 1; i <= n; i++) {
    a[i].nm = read();
    a[i].id = i;
  }
  sort(a + 1, a + 1 + n, cmp);
  int tot = 0;
  for (int i = 1 / sq; i <= n; i += sq) {
    int rr = max(1, min(n, i + sq - 1)), ll = max(i, 1);
    makeblock(1, ll, rr);
    B[i / sq] = tre[1];
    tot = i / sq;
  }
  for (int ii = 1; ii <= m; ii++) {
    s = read();
    o = read();
    int as = 0;
    for (int i = 0; i <= tot; i++) {
      int ll = lower_bound(B[i].b.begin(), B[i].b.end(), s) - B[i].b.begin();
      int rr =
          lower_bound(B[i].b.begin(), B[i].b.end(), o + 1) - B[i].b.begin() - 1;
      as = merg(as, B[i].mp[ll][rr]);
    }
    S[ii] = as;
  }
  printf("%d\n", cnt);
  for (int i = n + 1; i <= cnt; i++) {
    printf("%d %d\n", recx[i], recy[i]);
  }
  for (int i = 1; i <= m; i++) {
    printf("%d ", S[i]);
  }
  putchar('\n');
  return 0;
}