If \(\sum L>0\), then \(\sum X\geq \sum L>0\), so no \(X\) satisfies the conditions. Likewise, if \(\sum R<0\), then no \(X\) satisfies the condition because \(\sum X\leq \sum R <0\).

Otherwise, there always exists a conforming \(X\).

For each \(i=1,2,\ldots,N\), initialize with \(X_i=L_i\), then one can greedily determine the values for \(i=1,2,\ldots,N\) in order. Specifically, one can do as follows.

• For each \(i=1,2,\ldots\), repeat incrementing the value \(X_i\) by \(+1\) while \(X_i<R_i\) and \(\sum X<0\).

However, doing this naively does not finish within the time limit. The value to add to \(X_i\) can be found fast as follows.

• For \(i=1,2,\ldots\), let \(D_i=\min(R_i-L_i,-\sum X)\). Increment the value \(X_i\) by \(+D_i\).

The complexity is \(O(N)\).

N = int(input())
L, R = [0] * N, [0] * N
for i in range(N):
    L[i], R[i] = map(int, input().split())

if sum(L) > 0 or sum(R) < 0:
    print("No")
    exit()

X = L.copy()
sumX = sum(X)
for i in range(N):
    d = min(R[i] - L[i], -sumX)
    sumX += d
    X[i] += d

print("Yes")
print(*X)