DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
const int MAXN = 200000;
using namespace std;
struct node {
  long long sum;
  int col;
  long long lazy;
};
node tr[4 * MAXN], dont;
int n, m;
void push(int l, int r, int ind) {
  tr[ind].sum += (r - l + 1) * 1ll * tr[ind].lazy;
  if (l != r) {
    tr[ind * 2 + 1].lazy += (long long)tr[ind].lazy;
    tr[ind * 2 + 2].lazy += (long long)tr[ind].lazy;
    if (tr[ind].col != -1) {
      tr[ind * 2 + 1].col = tr[ind].col;
      tr[ind * 2 + 2].col = tr[ind].col;
    }
  }
  tr[ind].lazy = 0;
}
node mrg(node l, node r) {
  node temp;
  if (l.col == -17) return r;
  if (r.col == -17) return l;
  if (l.col == r.col) {
    temp.lazy = 0ll;
    temp.sum = (long long)l.sum + r.sum;
    temp.col = l.col;
  } else {
    temp.lazy = 0ll;
    temp.sum = (long long)l.sum + r.sum;
    temp.col = -1;
  }
  return temp;
}
void build(int l, int r, int ind) {
  if (l > r) return;
  if (l == r) {
    tr[ind].lazy = 0ll;
    tr[ind].sum = 0ll;
    tr[ind].col = l;
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, ind * 2 + 1);
  build(mid + 1, r, ind * 2 + 2);
  tr[ind] = mrg(tr[ind * 2 + 1], tr[ind * 2 + 2]);
}
void update(int l, int r, int ql, int qr, int val, int ind) {
  push(l, r, ind);
  if (l > r || l > qr || r < ql) return;
  if (ql <= l && r <= qr && tr[ind].col != -1) {
    tr[ind].lazy += (long long)abs(val - tr[ind].col);
    tr[ind].col = val;
    push(l, r, ind);
    return;
  }
  int mid = (l + r) / 2;
  update(l, mid, ql, qr, val, ind * 2 + 1);
  update(mid + 1, r, ql, qr, val, ind * 2 + 2);
  tr[ind] = mrg(tr[ind * 2 + 1], tr[ind * 2 + 2]);
}
node query(int l, int r, int ql, int qr, int ind) {
  push(l, r, ind);
  if (l > r || l > qr || r < ql) return dont;
  if (ql <= l && r <= qr) return tr[ind];
  int mid = (l + r) / 2;
  return mrg(query(l, mid, ql, qr, ind * 2 + 1),
             query(mid + 1, r, ql, qr, ind * 2 + 2));
}
void mainp() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  build(0, n, 0);
  dont.sum = 0ll;
  dont.lazy = 0ll;
  dont.col = -17;
  for (int i = 0; i < m; i++) {
    int type;
    cin >> type;
    if (type == 1) {
      int l, r, x;
      cin >> l >> r >> x;
      update(0, n, l, r, x, 0);
    } else {
      int l, r;
      cin >> l >> r;
      cout << query(0, n, l, r, 0).sum << '\n';
    }
  }
}
int main() {
  mainp();
  return 0;
}
