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;
const long long MOD = 1000 * 1000 * 1000 + 7;
const long long NMAX = 1000 * 1000 + 11;
long long col[NMAX], lazy[NMAX], ans[NMAX];
long long n, m, p, q, k;
void build(long long l, long long r, long long p) {
  if (l == r) {
    col[p] = l + 1;
    return;
  }
  build(l, (l + r) / 2, 2 * p);
  build((l + r) / 2 + 1, r, 2 * p + 1);
  return;
}
void check(long long p, long long l, long long r) {
  if (!lazy[p]) return;
  col[2 * p] = col[p];
  col[2 * p + 1] = col[p];
  long long mid = (l + r) >> 1LL;
  ans[2 * p] += (lazy[p]) * (mid - l + 1);
  ans[2 * p + 1] += (lazy[p]) * (r - mid);
  if (l != r) {
    lazy[2 * p] += lazy[p];
    lazy[2 * p + 1] += lazy[p];
  }
  lazy[p] = 0;
  return;
}
void update(long long p, long long l, long long r, long long qs, long long qe,
            long long x) {
  check(p, l, r);
  if (l > qe || r < qs) return;
  if (qs <= l and r <= qe) {
    if (col[p]) {
      ans[p] += abs(col[p] - x) * (r - l + 1);
      lazy[p] += abs(col[p] - x);
      col[p] = x;
    } else {
      long long mid = (l + r) / 2;
      update(2 * p, l, mid, qs, qe, x);
      update(2 * p + 1, mid + 1, r, qs, qe, x);
      ans[p] = ans[2 * p] + ans[2 * p + 1];
      col[p] = x;
    }
    return;
  }
  long long mid = (l + r) / 2;
  update(p * 2, l, mid, qs, qe, x);
  update(p * 2 + 1, mid + 1, r, qs, qe, x);
  ans[p] = ans[p * 2] + ans[p * 2 + 1];
  col[p] = col[p * 2] == col[p * 2 + 1] ? col[p * 2] : 0;
  return;
}
long long query(long long p, long long l, long long r, long long qs,
                long long qe) {
  check(p, l, r);
  if (l > qe || qs > r) return 0;
  if (qs <= l and r <= qe) return ans[p];
  long long mid = (l + r) / 2;
  return query(p * 2, l, mid, qs, qe) + query(p * 2 + 1, mid + 1, r, qs, qe);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  build(0, n - 1, 1);
  long long type, l, r, x;
  while (m--) {
    cin >> type;
    if (type == 2) {
      cin >> l >> r;
      cout << query(1, 0, n - 1, l - 1, r - 1) << "\n";
    } else {
      cin >> l >> r >> x;
      update(1, 0, n - 1, l - 1, r - 1, x);
    }
  }
  return 0;
}
