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 int maxn = 100010;
const int maxm = 400010;
const int INF = 0x3f3f3f3f;
const long long LINF = 3e17 + 1;
const int mod = 1e9 + 7;
int n, r, c, m;
inline int read() {
  register int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return (f == 1) ? x : -x;
}
struct seg_tree {
  long long l, r;
  long long sum;
  long long num;
  long long lz;
} tr[maxn << 2];
void pushup(long long o) {
  tr[o].sum = tr[(o << 1)].sum + tr[(o << 1 | 1)].sum;
  if (tr[(o << 1)].num == tr[(o << 1 | 1)].num)
    tr[o].num = tr[(o << 1)].num;
  else
    tr[o].num = 0;
}
void pushdown(long long o) {
  if (tr[o].lz) {
    long long mid = (tr[o].l + tr[o].r) >> 1;
    tr[(o << 1)].sum += (mid - tr[o].l + 1) * tr[o].lz;
    tr[(o << 1 | 1)].sum += (tr[o].r - mid) * tr[o].lz;
    tr[(o << 1)].num = tr[(o << 1 | 1)].num = tr[o].num;
    tr[(o << 1)].lz += tr[o].lz;
    tr[(o << 1 | 1)].lz += tr[o].lz;
    tr[o].lz = 0;
  }
}
void build(long long o, long long l, long long r) {
  tr[o].l = l, tr[o].r = r, tr[o].lz = 0, tr[o].sum = 0;
  if (l == r) {
    tr[o].num = l;
    return;
  }
  long long mid = (l + r) >> 1;
  build((o << 1), l, mid);
  build((o << 1 | 1), mid + 1, r);
  pushup(o);
}
void update(long long o, long long l, long long r, long long ql, long long qr,
            long long val) {
  if (ql <= l && r <= qr && tr[o].num) {
    tr[o].sum += (r - l + 1) * abs(val - tr[o].num);
    tr[o].lz += abs(val - tr[o].num);
    tr[o].num = val;
    return;
  } else if (l == r) {
    tr[o].sum += abs(val - tr[o].num);
    tr[o].num = val;
    return;
  }
  long long mid = (l + r) >> 1;
  pushdown(o);
  if (ql <= mid) update((o << 1), l, mid, ql, qr, val);
  if (mid < qr) update((o << 1 | 1), mid + 1, r, ql, qr, val);
  pushup(o);
}
long long query(long long o, long long l, long long r, long long ql,
                long long qr) {
  if (ql <= l && r <= qr) {
    return tr[o].sum;
  }
  pushdown(o);
  long long mid = (l + r) >> 1;
  long long res = 0;
  if (ql <= mid) res += query((o << 1), l, mid, ql, qr);
  if (mid < qr) res += query((o << 1 | 1), mid + 1, r, ql, qr);
  return res;
}
int main() {
  ;
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n >> m;
  build(1, 1, n);
  int opt, l, r;
  long long v;
  while (m--) {
    cin >> opt >> l >> r;
    if (opt == 1) {
      cin >> v;
      update(1, 1, n, l, r, v);
    } else {
      cout << query(1, 1, n, l, r) << endl;
    }
  }
  return 0;
}
