satisfy the momentum conservation. The events remain if the adjusted total momentum is less than 0.017 GeV , i.e. Where f and f are supposed to be a pair of opposite fractional charge particles, p ( f ) , p ( f ) , p ( e $^{+}$) and p ( e − ) are measured momenta for the four tracks. If more than one pair of opposite charge tracks satisfy the requirement, the combination with minimal adjusted total momentum is chosen. The unadjusted pair is supposed to be a pair of e $^{+}$e − and is required to have invariant mass | m$_{e}$ $^{+}$e − − 3 . 097 | < 0 . 100 GeV . The electron (positron) track is required to have zero hit in the muon detector. The number of events for the remaining events as well as the measured charge q and mass m$_{f}$ ( m$_{f}$ ) for the signal channels are listed in table 1 and table 2. The efficiency to select the first channel is larger than that of the second channel, because the d and d $^{′}$s bending radiuses in the detector are two time those of the u and u if they carry the same real momentum and the d and d arrive at the subdetector of the flight time easier. The Fig. 1 shows the distribution of the charge q$_{d}$ and Fig. 2 shows the scatter of the mass m$_{d}$ ( m$_{d}$ ) versus the charge q$_{d}$ . It can be seen from the two plots that the measured charge q$_{d}$ concentrates at 0.333 and mass m$_{d}$ concentrates 0.010 GeV for the channel ψ (2 S ) → ddJ/ψ with the input mass m$_{d}$ = 0 . 005 GeV . 22 events remain from the background sample, which are also plotted in Fig.1 and Fig. 2. There is only one event in the d charge window | q$_{d}$ − 0 . 333 | < 0 . 040 and 4 events in the u charge window | q$_{u}$ − 0 . 666 | < 0 . 040 . In summary, an inclusive sample can be divided into exclusive channels n$_{1}$πs + n$_{2}$Ks + n$_{3}$ps + n$_{4}$e $^{+}$e − s + n$_{5}$µ $^{+}$µ − s + n$_{6}$γs + n$_{7}$fs, where f denotes a fractional charge particle. For a fractional charge track, its nominal momentum, derived by assuming its charge equal to 1 when it is reconstructed, increases by a factor 1 /q com- pared with its real momentum. Then the nominal total momentum and energy for the event 6 Instead to calculate the f and f masses, the masses squared are derived to be m 2 f = q $^{2}$p$^{2}$( f )(1 − v $^{2}$( f )) /v $^{2}$( f ) , m$_{f}$ 2 = q $^{2}$p$^{2}$( f )(1 − v $^{2}$( f )) /v $^{2}$( f ) , where v ( f ) and v ( f ) are the measured velocities and often larger than 1 due to measurement errors. Then the total energy is required to be | E (tot) − 3 . 686 | < 0 . 15 GeV, where E (tot) = √ ( qp ) $^{2}$( f ) + m 2 f + √ ( qp ) $^{2}$( f ) + m 2 f + √ ( p ) $^{2}$( e $^{+}$) + m 2 e + + √ ( p ) $^{2}$( e − ) + m 2 e − . P$_{r}$ (tot) = √ ∑ i = x, y, z ( qp$_{i}$ ( f ) + qp$_{i}$ ( f ) + p$_{i}$ ( e $^{+}$) + p$_{i}$ ( e − )) 2 < 0 . 017 GeV,