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>
using namespace std;
long long int pw(long long int a, long long int b, long long int mod) {
  if (!b) return 1;
  if (b & 1) return a * pw(a * a % mod, b / 2, mod) % mod;
  return pw(a * a % mod, b / 2, mod) % mod;
}
const long long int MAXN = 1e5 + 10;
const long long int INF = 8e18;
const long long int MOD = 1e9 + 7;
long long int n, q, seg[MAXN << 2][3], lazy[MAXN << 2][2];
void build(int id, int l, int r) {
  if (r - l == 1) {
    seg[id][1] = seg[id][2] = r;
    return;
  }
  int mid = (l + r) >> 1;
  build(2 * id, l, mid);
  build(2 * id + 1, mid, r);
  seg[id][1] = min(seg[2 * id][1], seg[2 * id + 1][1]);
  seg[id][2] = max(seg[2 * id][2], seg[2 * id + 1][2]);
}
void shift(int id, int l, int r) {
  if (lazy[id][0] == 0) return;
  seg[id][0] += lazy[id][1] * (r - l);
  seg[id][1] = seg[id][2] = lazy[id][0];
  if (r - l > 1) {
    lazy[2 * id][1] += lazy[id][1];
    lazy[2 * id + 1][1] += lazy[id][1];
    lazy[2 * id][0] = lazy[2 * id + 1][0] = lazy[id][0];
  }
  lazy[id][0] = lazy[id][1] = 0;
}
void update(int id, int l, int r, int ql, int qr, long long int x) {
  shift(id, l, r);
  if (qr <= l || r <= ql) return;
  if (ql <= l && r <= qr && seg[id][1] == seg[id][2]) {
    lazy[id][0] = x;
    lazy[id][1] = abs(x - seg[id][1]);
    return shift(id, l, r);
  }
  int mid = (l + r) >> 1;
  update(2 * id, l, mid, ql, qr, x);
  update(2 * id + 1, mid, r, ql, qr, x);
  seg[id][0] = seg[2 * id][0] + seg[2 * id + 1][0];
  seg[id][1] = min(seg[2 * id][1], seg[2 * id + 1][1]);
  seg[id][2] = max(seg[2 * id][2], seg[2 * id + 1][2]);
}
long long int get(int id, int l, int r, int ql, int qr) {
  shift(id, l, r);
  if (qr <= l || r <= ql) return 0;
  if (ql <= l && r <= qr) return seg[id][0];
  int mid = (l + r) >> 1;
  return get(2 * id, l, mid, ql, qr) + get(2 * id + 1, mid, r, ql, qr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> q;
  build(1, 0, n);
  while (q--) {
    int type, l, r, x;
    cin >> type >> l >> r;
    if (type == 1) {
      cin >> x;
      update(1, 0, n, l - 1, r, x);
    } else {
      cout << get(1, 0, n, l - 1, r) << '\n';
    }
  }
  return 0;
}
